# Structure of Atom

## Chemistry Class 11

### NCERT

1   (i) Calculate the number of electrons which will together weigh one gram.$\\$ (ii) Calculate the mass and charge of one mole of electrons.

6   Find energy of each of the photons which $\\$ (i) correspond to light of frequency $3× 10 ^{15} Hz.$$\\ (ii) have wavelength of 0.50 Å \\ ##### Solution : (i) Energy (E) of a photon is given by the expression,\\ E = hv \\ Where,\\ h = Planck’s constant = 6.626×10^{-31}Js \\ v = frequency of light = 3×10^{15} Hz \\ Substituting the values in the given expression of E: E = (6.626×10^{-34} ) (3×10 ^{15} ) \\ E = 1.988×10 ^{-18}J \\ (ii) Energy (E) of a photon having wavelength (\lambda ) is given by the expression, E =\dfrac{hc}{\lambda} \\ h = Planck’s constant = 6.626×10 ^{-34}Js \\ c = velocity of light in vacuum = 3×10^ 8 m/s Substituting the values in the given expression of E: \\ E=\dfrac{(6.626*10^{-34})(3*10^8)}{0.50*10^{-10}} \\ E=3.98*10^{-15}J \\ 7 Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10^{-10}s. ##### Solution : Frequency (v) of light =\dfrac{1}{\text{Period}} \\ =\dfrac{1}{2.0*10 ^{-10}s}=5.0*10^9 s^{-1} \\ Wavelength (\lambda ) of light =\dfrac{c}{v} \\ Where,\\ c = velocity of light in vacuum = 3×10^ 8 m/s \\ Substituting the value in the given expression of \lambda \\ \lambda =\dfrac{3*10^8}{5.0*10^9}=6.0*10^{-2}m \\ Wave number (\bar{ v} ) of light \dfrac{1}{\lambda }=\dfrac{1}{6.0*10^{-2}}=1.66*10^{1}m^{-1}=16.66 m 8 What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy? ##### Solution : Energy (E) of a photon = hv \\ Energy (E n ) of ‘n’ photons = nhv \\ \Rightarrow n=\dfrac{E_n\lambda }{hc} \\ Where,\\ \lambda =wavelength of light = 4000 pm = 4000*10^{-12}m \\ c = velocity of light in vacuum = 3 × 10^ 8 m/s \\ h = Planck’s constant = 6.626 × 10^{-34} Js \\ Substituting the values in the given expression of n: \\ n=\dfrac{(1)*(4000*10^{-12})}{(6.626 * 10^{-34})(3*10^8)}=2.012*10^{16} \\ Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012×10 ^{16 }. 9 A photon of wavelength 4 × 10^{-7} m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate\\ (i) the energy of the photon (eV), \\ (ii) the kinetic energy of the emission, and\\ (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10^{-19} J). ##### Solution : (i) Energy (E) of a photon = hv =\dfrac{hc}{\lambda } \\ Where, \\ h = Planck’s constant = 6.626 × 10 ^{-34}Js \\ c = velocity of light in vacuum = 3 × 10^8 m/s \\ \lambda = wavelength of photon = 4 × 10^{-7} m \\ Substituting the values in the given expression of E:\\ E=\dfrac{(6.626*10 ^{-34})(3*10^8)}{4*10^{-7}}=4.9695*10^{-19}J \\ Hence, the energy of the photon is 4.97×10^{-19} J. \\ (i) The kinetic energy of emission E_ k is given by \\ = hv - hv_ 0 \\ =( E - W ) eV \\ =(\dfrac{4.9695 * 10 ^{- 19}}{1.6020 * 10^{-19}}) eV-2.13eV \\ = (3.1020 - 2.13) eV \\ = 0.9720 eV \\ Hence, the kinetic energy of emission is 0.97 eV. \\ (ii) The velocity of a photoelectron (v) can be calculated by the expression,\\ \dfrac{1}{2}mv^2=hv=hv_o \\ \Rightarrow v=\sqrt{2(hv-hv_o)}{m} \\ Where, ( hv -hv _0) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron.\\ Substituting the values in the given expression of v: \\ v=\sqrt{\dfrac{2*(0.9720*1.6020*10^{-19})}{9.10939*10^{-31}kg}J} \\ =\sqrt{0.3418*10^{12}m^2s^{-2}} \\ v=5.84*10^5 ms^{-1} \\ Hence, the velocity of the photoelectron is 5.84 × 10 ^5 ms ^{-1}. 10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol ^{-1} ##### Solution : Energy of sodium (E) =\dfrac{N_Ahc}{\lambda } \\ =\dfrac{(6.023* 10^{23 }mol^{-1})(6.626*10^{-34}Js)(3*10^8 ms^{-1})}{242*10^{-9}m} \\ =4.947*10^5 J mol^{-1} \\ =494.7*10^3 J mol^{-1} \\ =494 kJ mol^{-1} 11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 \mu m. Calculate the rate of emission of quanta per second. ##### Solution : Power of bulb, P = 25 Watt = 25 Js^{-1}$$\\$ Energy of one photon, E = hv =$\dfrac{hc}{\lambda}$$\\ Substituting the values in the given expression of E:\\ E=\dfrac{( 6.626 * 10 ^{- 34}) (3*10^8)}{( 0.57 * 10^{-6} )}=34.87*10^{-20} J\\ E=34.87*10^{-20}J$$\\$ Rate of emission of quanta per second$\\$ $=\dfrac{25}{34.87*10^{-20}}=7.169*10^{19}s^{-1}$

12   Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency ( $v _0$) and work function ( $w _0$ ) of the metal.

##### Solution :

Threshold wavelength of radian $(\lambda_ 0 )= 6800 Å = 6800 × 10^{-10} m$$\\ Threshold frequency (v _0)of the metal\\ \dfrac{c}{\lambda_0}=\dfrac{3*10^8ms^{-1}}{6.8*10^{-7}m}=4.41*10^{14}s^{-1} \\ Thus, the threshold frequency ( v_ 0 ) of the metal is 4.41*10^{14}s^{-1} \\ Hence, work function ( w_ 0 ) of the metal = hv_ 0$$\\$ $= (6.626×10^{-34}Js)(4.41*10^{14}s^{-1})\\ = 2.922×10^{-19}J$

13   What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

The $n_ i = 4$ to $n_ f = 2$ transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,$\\$ $E=2.18*10^{-18}[\dfrac{1}{n_1^2}-\dfrac{1}{n^2_f}]$ $\\$ Substituting the values in the given expression of E:$\\$ $E=2.18*10^{-18}[\dfrac{1}{4^2}-\dfrac{1}{2^2_f}]$ $\\$ $=2.18*10^{-18}[\dfrac{1-4}{16}]$ $\\$ $=2.18*10^{-18}[\dfrac{-3}{16}]$ $\\$ $E=-(4.0875 × 10^{-19} J)$$\\ The negative sign indicates the energy of emission.\\ Wavelength of light emitted (\lambda )=\dfrac{hc}{E} \\ Since E=\dfrac{hc}{\lambda} \\ Substituting the values in the given expression of \lambda \\ \lambda=\dfrac{( 6.626 * 10^{-34} )( 3 * 10 ^8)}{4.0875 *10 ^{- 19}}\\ =4.8631 * 10 ^{- 7} m\\ = 486.3 * 10 ^{- 9}m\\ = 486 nm 14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit). ##### Solution : The expression of energy is given by,\\ E_n=\dfrac{-(2.18 * 10 ^{- 18})Z^2}{n^2} \\ Where,\\ Z = atomic number of the atom\\ n = principal quantum number\\ For ionization from n _1 = 5 to n_ 2 =\infty , \Delta E = E _{\infty}- E _5\\ =[(\dfrac{(-2.18*10^{-18}J)(1)^2}{(\infty)^2})-(\dfrac{(-2.18*10^{-18}J)(1)^2}{(5)^2})]J\\ (2.18*10^{-18})J \\ Hence, the energy required for ionization from n = 5 to n =\infty is 8.72 × 10^{-20}JEnergy required for n_ 1 =1 to n=\infty,\\ \Delta E=E_{\infty}-E_5\\ =[(\dfrac{(-2.18*10^{-18}J)(1)^2}{(\infty)^2})-(\dfrac{(-2.18*10^{-18}J)(1)^2}{(1)^2})]J\\ =(- 2.18* 10 ^{- 18} J )( 1 - 0 )\\ (2.18*10^{-18})J \\ Hence, less energy is required to ionize an electron in the 5^{ th} orbital of hydrogen atom as compared to that in the ground state. 15 What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state? ##### Solution : When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible: Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum. The number of spectral lines produced when an electron in the n th level drops down to the ground state is given by. Given, =\dfrac{n(n-1)}{2}\\ n=6 \\ Number of spectral lines =\dfrac{6(6-1)}{2}=15 16 (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10^{-18}J atom^{-1}.What is the energy associated with the fifth orbit?\\ (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.\\ ##### Solution : (i) Energy associated with the fifth orbit of hydrogen atom is calculated as:\\ E_5=\dfrac{-(2.18*10^{-18})}{(5)^2}=\dfrac{-2.18*10^{-18}}{25}\\ E_5=-8.72*10^{-20}J \\ (ii) Radius of Bohr’s n^{ th} orbit for hydrogen atom is given by,\\ r_n=(0.0529 nm)n^2 \\ For,\\ n = 5\\ r _5 = (0.0529 nm) (5)^ 2\\ r _5 = 1.3225 nm$$\\$

17   Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

##### Solution :

For the Balmer series, $n _i = 2.$ Thus, the expression of wave number ($\bar{v}$) is given by,$\\$ $\bar{v}=(\dfrac{1}{(2)^2}-\dfrac{1}{n^2_f})(1.097*10^7 m^{-1})$ $\\$ Wave number $(\bar{v})$ is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, $(\bar{v})$has to be the smallest.$\\$ For $(\bar{v})$ to be minimum, $n_ f$ should be minimum. For the Balmer series, a transition from $n_ i = 2$ to $n_ f = 3$ is allowed. Hence, taking $n_ f = 3$, we get:$\\$ $\bar{v}=1.5236*10^6 m^{-1}\\ \bar{v}=(1.097 * 10 7 m ^{- 1})(\dfrac{1}{(2)^2}-\dfrac{1}{3^2})\\ \bar{v}=(1.097 * 10 7 m ^{- 1})(\dfrac{1}{4}-\dfrac{1}{9})\\ \bar{v}=(1.097*10^7m^{-1})(\dfrac{9-4}{36})\\ \bar{v}=(1.097*10^7m^{-1})(\dfrac{5}{36})\\ \bar{v}=1.5236 * 10 ^6 m ^{- 1}$

18   What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is $–2.18 * 10 ^{- 11}$ergs .

##### Solution :

Energy (E) of the n th Bohr orbit of an atom is given by,$\\$ $E_n=\dfrac{-(2.18*10^{-18})Z^2}{n^2}$ $\\$ Where,$\\$ Z = atomic number of the atom$\\$ Ground state energy =$-2.18*10^{-11}ergs\\ =-2.18*10^{-11}*10^{-7}J\\ =-2.18*10^{-18}J$ $\\$ Energy required to shift the electron from $n = 1$ to $n = 5$is given as:$\\$ $\Delta E=E_5-E_1\\ =\dfrac{-(2.18*10^{-18})(1)^2}{(5)^2}-(-2.18*10^{-18})\\ =(2.18*10^{-18})[1-\dfrac{1}{25}]\\ =(2.18*10^{-18})(24/25)=2.0928*10^{-18}$ $\\$ Weight of emitted light =$\dfrac{hc}{E}\\ =\dfrac{(6.626*10^{-34}){3*10^8}}{(2.0928*10^{-18})}\\ =9.498*10^{-8}m$

19   The electron energy in hydrogen atom is given by $En = (–2.18 × 10^{-18} )/n^2$ J.Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

##### Solution :

$E_n=\dfrac{2.18*10^{-18}}{n^2}J$ $\\$ Given $\\$ $\Delta E=E_{\infty}-E_2\\ =[(\dfrac{-2.18*1110^{-18}}{(\infty)^2})-(\dfrac{-2.18*1110^{-18}}{(2)^2})]J\\ =(\dfrac{2.18*10^{-18}}{4}-0)J$ $\\$ Energy required for ionization from $n = 2$ is given by,$\\$ $=0.545*10^{-18}J\\ \Delta E=5.45*10^{-19}J\\ \lambda=\dfrac{hc}{\Delta E}$ $\\$ Here, $\lambda$ is the longest wavelength causing the transition.$\\$ $\lambda =\dfrac{(6.626*10^{-34})(3*10^8)}{5.45*10^{-19}}=3.647*10^{-7}m\\ =3647*10^{-10}m\\ =3647 Å$

20   Calculate the wavelength of an electron moving with a velocity of $2.05 × 10^7 ms ^{-1}$.

##### Solution :

According to de Broglie’s equation,$\\$ $\lambda =\dfrac{h}{mv}$ $\\$ Where,$\\$ $\lambda$= wavelength of moving particle$\\$ m = mass of particle, v = velocity of particle, h = Planck’s constant$\\$ Substituting the values in the expression of $\lambda$:$\\$ $\lambda=\dfrac{6.626*10^{-34}Js}{(9.10039*10^{-31}kg)(2.05*10^7ms^{-1})}\\ \lambda=3.548*10^{-11}m$ $\\$ Hence, the wavelength of the electron moving with a velocity of$2.05*10^7ms^{-1}$ is $3.548*10^{-11}m$

21   The mass of an electron is $9.1 × 10^{31}$ kg If its K.E. is $3.0 × 10 ^{–25} J ,$calculate its wavelength.

##### Solution :

From de Broglie’s equation,$\\$ $\lambda=\dfrac{h}{mv}$ $\\$ Given, Kinetic energy (K.E.) of the electron = $3.0 × 10 ^{–25} J$$\\ Since K.E. =\dfrac{1}{2}mv^2 \\ Velocity ( v )=\sqrt{\dfrac{2K.E.}{m}}\\ =\sqrt{\dfrac{2(3.0*10^{-25}J)}{9.10939*10^{-31}kg}}\\ =\sqrt{6.5866 * 10 ^4}\\ V=811.579ms^{-1}$$\\$ Substituting the value in the expression of $\lambda$:$\\$ $\lambda =\dfrac{6.626*10^{-34}Js}{(9.10939*10^{-31}kg)(811.579ms^{-1})}\\ \lambda=8.9625*10^{-7}m$ $\\$ Hence, the wavelength of the electron is$8.9625×10 ^{–7} m.$

22   Which of the following are isoelectronic species i.e., those having the same number of electrons? $Na ^+ , K ^+ , Mg^{ 2+}, Ca ^{2+} , S ^{2-} , Ar$

##### Solution :

Isoelectronic species have the same number of electrons.$\\$ Number of electrons in sodium $(Na) = 11$$\\ Number of electrons in (Na ^+ ) = 10$$\\$ A positive charge denotes the loss of an electron. Similarly,$\\$ Number of electrons in $K^ + = 18$$\\ Number of electrons in Mg^{ 2+} = 10$$\\$ Number of electrons in$Ca^{ 2+} = 18$$\\ A negative charge denotes the gain of an electron by a species.\\ Number of electrons in sulphur (S) = 16$$\\$ $\therefore$Number of electrons in $S 2- = 18$$\\ Number of electrons in argon (Ar) = 18$$\\$ Hence, the following are isoelectronic species:$\\$ (1) $Na^ +$ and $Mg ^{2+}$ (10 electrons each)$\\$ (2) $K ^+$, $Ca ^{2+} , S ^{2-}$ and Ar (18 electrons each).$\\$

23   (i) Write the electronic configurations of the following ions: $(a) H ^– (b) Na^+ (c) O^{2–} (d) F ^–$$\\ (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s^ 1 (b) 2p^ 3 \ \ and\ \ (c) 3p^ 5 ?$$\\$ (iii) Which atoms are indicated by the following configurations? $(a) [He] 2s^ 1\ \ (b) [Ne] 3s ^2 3p^ 3 \ \ (c) [Ar] 4s^ 2 3d ^1 .$

##### Solution :

(i) (a) $H^ –$ ion$\\$ The electronic configuration of H atom is 1s$^ 1$.$\\$ A negative charge on the species indicates the gain of an electron by it.$\\$ $\therefore$ Electronic configuration of H $^– = 1s^ 2$$\\ (b) Na^+ ion\\ The electronic configuration of Na atom is 1s^ 2 \ 2s ^2\ 2p ^6\ 3s ^1 .$$\\$ A positive charge on the species indicates the loss of an electron by it.$\\$ $\therefore$ Electronic configuration of Na$^+ = 1s ^2\ 2s ^2\ 2p ^6\ 3s^ 0 \ \ or \ \ 1s ^2\ 2s ^2\ 2p ^6$$\\ (c) O2^– ion\\ The electronic configuration of 0 atom is 1s ^2\ 2s ^2\ 2p ^4 .\\ A dinegative charge on the species indicates that two electrons are gained by it.\\ \therefore Electronic configuration of O ^{2 -} ion = 1s^ 2\ 2s ^2 \ p ^6$$\\$ (d) F$^ –$ ion$\\$ The electronic configuration of F atom is 1s 2 2s 2 2p 5 .$\\$ A negative charge on the species indicates the gain of an electron by it.$\\$ $\therefore$ Electron configuration of F – ion = 1s 2 2s 2 2p 6$\\$

##### Solution :

(i) n = 3 (Given)$\\$ For a given value of n, l can have values from 0 to (n – 1).$\\$ $\therefore$ For n = 3$\\$ l = 0, 1, 2$\\$ For a given value of l, m l can have (2l + 1) values.$\\$ For l = 0, m = 0$\\$ l = 1, m = –1, 0, 1$\\$ l = 2, m = –2, –1, 0, 1, 2$\\$ $\therefore$ For n = 3$\\$ l = 0, 1, 2$\\$ m 0 = 0$\\$ m 1 = –1, 0, 1$\\$ m 2 = –2, –1, 0, 1, 2$\\$ (ii) For 3d orbital, l = 2.$\\$ For a given value of l, m l can have (2l + 1) values i.e., 5 values.$\\$ $\therefore$ For l = 2$\\$ m 2 = –2, –1, 0, 1, 2$\\$ (iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist. For p-orbital, l = 1. For a given value of n, l can have values from zero to (n – 1).$\\$ $\therefore$ For l is equal to 1, the minimum value of n is 2.$\\$ Similarly,$\\$ For f-orbital, l = 4.$\\$ For l = 4, the minimum value of n is 5. Hence, 1p and 3f do not exist.$\\$

29   Using s, p, d notations, describe the orbital with the following quantum numbers.$\\$ (a) n = 1, l = 0;$\\$ (b) n = 3; l =1$\\$ (c) n = 4; l = 2;$\\$ (d) n = 4; l =3.$\\$

##### Solution :

(a) n = 1, l = 0 (Given) The orbital is 1s.$\\$ (b) For n = 3 and l = 1 The orbital is 3p.$\\$ (c) For n = 4 and l = 2 The orbital is 4d.$\\$ (d) For n = 4 and l = 3 The orbital is 4f.$\\$

30   Explain, giving reasons, which of the following sets of quantum numbers are not$\\$ $\begin{array}{|c|c|} \hline A&n = 0&l = 0&m_ l = 0&m_s=+\frac{1}{2}\\ \hline B& n = 1 &l = 0 &m_ l = 0 &m _s=-\frac{1}{2}\\ \hline C &n = 1& l = 1 &m _l = 0 &m _s=+\frac{1}{2}\\ \hline D& n = 2& l = 1 &m_ l = 0&m _s=-\frac{1}{2}\\ \hline E& n = 3& l = 3 &m_ l = – 3 &m _s=+\frac{1}{2}\\ \hline F &n = 3 &l = 1 &m_ l = 0 &m _s=+\frac{1}{2}\\ \hline \hline \end{array}$

##### Solution :

(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.$\\$ (b) The given set of quantum numbers is possible.$\\$ (c) The given set of quantum numbers is not possible. For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.$\\$ (d) The given set of quantum numbers is possible.$\\$ (e) The given set of quantum numbers is not possible. For n = 3,$\\$ $l = 0 to (3 – 1)$$\\ l = 0 to 2 i.e., 0, 1, 2$$\\$ (f) The given set of quantum numbers is possible.$\\$

31   How many electrons in an atom may have the following quantum numbers?$\\$ (a) $n = 4,$ and $m _s=-\dfrac{1}{2} $$\\ (b) n = 3, l = 0 ##### Solution : (a) Total number of electrons in an atom for a value of n = 2n^ 2$$\\$ $\therefore$ For $n = 4,$$\\ Total number of electrons = 2 (4)^ 2$$\\$ =$32$$\\ The given element has a fully filled orbital as 1s ^2\ 2s ^2\ 2p ^6\ 3s ^2\ 3p ^6\ 4s^ 2\ 3d^{ 10 }.\\ Hence, all the electrons are paired.\\ \therefore Number of electrons (having n = 4 and m _s=-\frac{1}{2} ) = 16 \\ (b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.\\ 32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit. ##### Solution : Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:\\ mvr=n \dfrac{h}{2 \pi}........(1)$$\\$ Where,$\\$ n = 1, 2, 3, ...$\\$ According to de Broglie’s equation:$\\$ $\lambda= \dfrac{h}{mv}........(1)\\ or \ \ \ mv=\dfrac{h}{\lambda}..........(2)$$\\ Substituting the value of ‘mv’ from expression (2) in expression (1):\\ \dfrac{h}{\lambda}=n \dfrac{h}{2 \pi}......(3)$$\\$ Since ‘$2\pi r$’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

33   What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition $n = 4$ to $n = 2$ of He$^+$ spectrum?

##### Solution :

For He$^+$ ion, the wave number ( $\bar{v}$ ) associated with the Balmer transition, $n = 4$ to $n = 2$ is given by:$\\$ $\bar{v}=\dfrac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ $\\$ Where,$\\$ $n_1=2\\ n_2=4$ $\\$ Z=atomic number of helium$\\$ $\bar{v}=\dfrac{1}{\lambda}=R(2)^2(\dfrac{1}{4}-\dfrac{1}{16})\\ =4R(\dfrac{4-1}{16})\\ \bar{v}=\dfrac{1}{\lambda}=\dfrac{3R}{4}\\ = \lambda=\dfrac{4}{3R}$ $\\$ According to the question, the desired transition for hydrogen will have the same wavelength as that of He$^+$.$\\$ $R(1)^2(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})=\dfrac{3R}{4}\\ (\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})=\dfrac{3}{4}.....(1)$ $\\$ By hit and trail method, the equality given by equation (1) is true only when$\\$ $n_ 1 = 1$ and $n _2 = 2$.$\\$ The transition for $n_ 2 = 2$ to $n = 1$ in hydrogen spectrum would have the same wavelength as Balmer transition $n = 4$ to $n = 2$ of He$^+$ spectrum.

34   Calculate the energy required for the process $He^+_{( g )} \to He ^{2+} _ {(g)} +e^-$ $\\$ The ionization energy for the H atom in the ground state is $2.18×10^{ –18} J$atom$^{-1}$

##### Solution :

Energy associated with hydrogen-like species is given by,$\\$ $E_n=-2.18*1111110^{-18}(\dfrac{Z^2}{n^2})J$ $\\$ For ground state of hydrogen atom,$\\$ $\Delta E=E_{\infty}=E_1\\ =0-[-2.18*10^{-18(\dfrac{1^2}{1^2})}]J\\ \Delta E=2.18*10^{-28}J$ $\\$ For the given process,$He^+_{(g)}\to He^{2+}_{(g)}+e^-$ $\\$ An electron is removed from $n = 1$ to $n = \infty.\\ \Delta E=E_{\infty}-E_1\\ =0-[-2.18*10^{-18}(\dfrac{(2)^2}{(1)^2})]J\\ \Delta E=8.72*10^{-18}J$ $\\$ The energy required for the process $8.72 * 10 ^{- 18} J$

35   If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

##### Solution :

$1 m = 100 cm$$\\ 1 cm = 10 ^{–2} m$$\\$ Length of the scale =$20 cm$$\\ = 20×10 ^{–2} m$$\\$ Diameter of a carbon atom = $0.15 nm$$\\ = 0.15×10 ^{–9} m$$\\$ One carbon atom occupies $0.15×10^{ –9} m$.$\\$ Number of carbon atoms that can be placed in a straight line$\\$ $=\dfrac{20*10^{-2}m}{0.15*10^{-9}m}\\ =133.33*10^7\\ =1.33*10^9$

36   $2×10^ 8$ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is$2.4 cm$.

Length of the given arrangement = 2.4 cm$\\$ Number of carbon atoms present = $2×10^ 8$$\\ \therefore Diameter of carbon atom\\ =\dfrac{2.4*10^{-2}m}{2*10^8m}\\ =1.2*10^{-16}m \\ Radius of carbon atom =\dfrac{\text{Diameter}}{2}\\ =\dfrac{1.2*10^{-10}m}{2}\\ =6.0*10^{-11}m \\ 37 The diameter of zinc atom is 2.6 A . Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. ##### Solution : (a) Radius of zinc atom =\dfrac{\text{Diameter}}{2}\\ =\dfrac{2.6A}{2}\\ =1.3*10^{-10}m\\ =130*10^{-12}m\\ 130 pm \\ (b) Length of the arrangement = 1.6 cm\\ =1.6*10^{-2}m \\ Diameter of zinc atom = 2.6×10^{ –10} m Number of zinc atoms present in the arrangement\\ =\dfrac{1.6*10^{-2}m}{2.6*10^{-10}m}\\ =0.6153*10^8m\\ =6.153*10^7 38 A certain particle carries 2.5×10^{ –16}C of static electric charge. Calculate the number of electrons present in it. ##### Solution : Charge on one electron = 1.6022×10 ^{–19} C$$\\$ =$1.6022×10 ^{–19}C$ charge is carried by 1 electron.$\\$ Number of electrons carrying a charge of $2.5×10^{ –16} C$ $\\$ $=1(\dfrac{2.5*10^{-16}C}{1.6022*10^{-19}})\\ =1.560*10^3C\\ =1560C$ $\\$

39   In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is $–1.282×10^{ –18}C$, calculate the number of electrons present on it.

##### Solution :

Charge on the oil drop = $1.282×10 ^{–18} C$$\\ Charge on one electron = 1.6022×10 ^{–19}C$$\\$ $\therefore$ Number of electrons present on the oil drop$\\$ $=\dfrac{1.282*10^{-18}C}{1.6022*10^{-19}C}\\ =0.8001*10^1\\ 8.0$

40   In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the $\alpha$-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

##### Solution :

A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms. Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of $\alpha$-particles (positively charged).

41   Symbols $^{79}_{35}Br$ and $^{79 }Br$ can be written, whereas symbols $^{35}_{ 79} Br$ and $^{35} Br$ are not acceptable. Answer briefly.

52   Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.$\\$$\begin{array}{|c|c|}\hline \lambda(nm)&500&450&400\\ \hline v*10^{-5}(cm s^{-1})&2.55&4.35&5.35 \\ \hline \end{array} ##### Solution : (a) Assuming the threshold wavelength to be =\lambda_0 nm =\lambda_ 0 * 10 ^{- 9} m , the kinetic energy of the radiation is given as:\\ h(v-v_0)=\dfrac{1}{2}mv^2 \\ Three different equalities can be formed by the given value as:\\ hc(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0})=\dfrac{1}{2}mv^2\\ hc(\dfrac{1}{500*10^9}-\dfrac{1}{\lambda_0*10^{-9}m})\\ =\dfrac{1}{2}m(2.55*10^{+5}*10^{-2}ms^{-1})\\ \dfrac{hc}{10^{-9}m}[\dfrac{1}{500}-\dfrac{1}{\lambda_0}]=\dfrac{1}{2}m(2.55*10^{+3}ms^{-1})^2 \ \ \ \ \ \ (1) \\ Similarly,\\ \dfrac{hc}{10^{-9}m}[\dfrac{1}{450}-\dfrac{1}{\lambda_0}]=\dfrac{1}{2}m(3.45*10^{+3}ms^{-1})^2 \ \ \ \ \ \ (2)\\ \dfrac{hc}{10^{-9}m}[\dfrac{1}{400}-\dfrac{1}{\lambda_0}]=\dfrac{1}{2}m(5.35*10^{+3}ms^{-1})^2 \ \ \ \ \ \ (3) \\ Dividing equation (3) by equation (1):\\ =\dfrac{[\dfrac{\lambda_0-400}{400 \lambda_0}]}{[\dfrac{\lambda_0-500}{500 \lambda_0}]}=\dfrac{(5.35*10^{+3}ms^{-1})^2}{(2.55*10^{+3}ms^{-1})^2}\\ \dfrac{5 \lambda_0-2000}{4 \lambda_0-2000}=(\dfrac{5.35}{2.55})^2=\dfrac{28.6225}{6.5025}\\ \dfrac{5 \lambda_0-2000}{4 \lambda_0-2000}=4.40177\\ 17.6070 \lambda_0=8803.537-2000\\ \lambda_0=\dfrac{6805.537}{12.607}\\ \lambda_0=539.8 nm\\ \lambda_0=540nm \\ Threshold wavelength \lambda_0 = 540 nm Note: part (b) of the question is not done due to the incorrect values of velocity given in the question.\\ 53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal. ##### Solution : From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W_ 0 ) of radiation and its kinetic energy (K.E.) i.e., E = W _0 + K.E.$$\\$ Energy of incident photon (E) =$\dfrac{hc}{\lambda}$$\\ Where,\\ c = velocity of radiation\\ h = Planck’s constant\\ \lambda = wavelength of radiation\\ Substituting the values in the given expression of E:\\ E=\dfrac{( 6.626 * 10^{-34}Js)(3.0 * 10^ 8 ms ^{- 1 })}{256.7 * 10 ^{- 9}m}\\ =7.744*10^{-19}J\\ \dfrac{7.44*10^{-19}}{1.602*10^{-19}eV}\\ E=4.83eV \\ The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,\\ K.E = 0.35 V\\ K.E = 0.35 eV\\ Work function, W 0 = E – K.E.\\ = 4.83 eV – 0.35 eV\\ = 4.48 eV\\ \dfrac{5 \lambda_0-2000}{4 \lambda_0-2000}=(\dfrac{5.35}{2.55})^2=\dfrac{28.6225}{6.5025}\\ \dfrac{5 \lambda_0-2000}{4 \lambda_0-2000}=4.40177\\ 17.6070 \lambda_0-5 \lambda_0=8803.537-2000\\ \lambda_0=\dfrac{6805.537}{12.607}\\ \lambda_0=539.8 nm\\ \lambda_0=540nm 54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5×10^ 7 ms ^{–1}, calculate the energy with which it is bound to the nucleus. ##### Solution : Energy of incident photon (E) is given by,\\ E=\dfrac{hc}{\lambda}\\ =\dfrac{(6.626 * 10^{-34} Js)(3.0 *10^8ms^{-1})}{(150*10^{-12}m)}\\ =1.3252*10^{-15}J\\ =13.252*10^{-16}J \\ Energy of the electron ejected (K.E)\\ =\dfrac{1}{2}m_ev^2\\ =\dfrac{1}{2}(9.10939*10^{31}kg)(1.5*10^7ms^{-1})^2\\ =10.2480*10^{-17}J\\ =1.025*10^{-16}J \\ Hence, the energy with which the electron is bound to the nucleus can be obtained as:\\ = E – K.E\\ = 13.252×10^{ –16 }J ^– 1.02×10 ^{–16} J\\ = 12.227×10 ^{–16} J\\ =\dfrac{12.227×10 ^{–16} }{1.602 * 10 ^{- 19}}eV\\ =7.6*10^3 eV\\ \dfrac{5 \lambda_0-2000}{4 \lambda_0-2000}=(\dfrac{5.35}{2.55})^2=\dfrac{28.6225}{6.5025}\\ \dfrac{5 \lambda_0-2000}{4 \lambda_0-2000}=4.40177\\ 17.6070 \lambda_0-5 \lambda_0=8803.537-2000\\ \lambda_0=\dfrac{6805.537}{12.607}\\ \lambda_0=539.8nm\\ \lambda=540nm 55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29×10 ^{15} (Hz) [1/3 ^2 - 1/n^ 2 ] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum. ##### Solution : Wavelength of transition = 1285 nm$$\\$ $= 1285×10^{ –9} m (Given)\\ v=3.29*10^{15}(\dfrac{1}{3^2-\dfrac{1}{n^2}})$ $\\$ (Given) $\\$ Since $v=\dfrac{c}{\lambda}\\ =\dfrac{3.0*10^8ms^{-1}}{1285*10^{-9}m}\\ v=2.33*10^{14}s^{-1}$ $\\$ Substituting the value of v in the given expression,$\\$ $3.29 * 10^{ 15}(\dfrac{1}{9}-\dfrac{1}{n^2})=2.33*10^{14}\\ \dfrac{1}{9}=\dfrac{1}{n^2}=\dfrac{2.33*10^{14}}{3.29*10^{15}}\\ \dfrac{1}{9}-0.7082*10^{-1}=\dfrac{1}{n^2}\\ \implies \dfrac{1}{n^2}=1.1*10^{-1}-0.7082*10^{-1}\\ \dfrac{1}{n^2}=4.029*106{-2}\\ n=\sqrt{\dfrac{1}{4.029*10^{-2}}}\\ n=4.98\\ n \approx 5$ $\\$ Hence, for the transition to be observed at $1285 nm, n = 5$. The spectrum lies in the infra-red region.

56   Calculate the wavelength for the emission transition if it starts from the orbit having radius $1.3225 nm$ and ends at $211.6 pm.$ Name the series to which this transition belongs and the region of the spectrum.

The radius of the $n^{ th}$ orbit of hydrogen-like particles is given by,$\\$ $r=\dfrac{0.529 n^ 2}{Z}Å\\ r=\dfrac{52.9 n^ 2}{Z}pm$ $\\$ For radius $( r_ 1 ) = 1.3225 nm\\ = 1.32225×10 ^{–9} m\\ = 1322.25×10^{ –12} m\\ = 1322.25 pm\\ n_1^2=\dfrac{r_1Z}{52.9}\\ n_2^1=\dfrac{1322.25Z}{52.9}$$\\ Similarly,\\ n_2^2=\dfrac{211.6Z}{52.9}\\ \dfrac{n_1^2}{n_2^2}=\dfrac{1322.5}{211.6}\\ \dfrac{n_1^2}{n_2^2}=6.25\\ \dfrac{n_1^2}{n_2^2}=2.5\\ \dfrac{n_1^2}{n_2^2}=\dfrac{25}{10}=\dfrac{5}{2}\\ \implies n_1=5 and n_2=2 \\ Thus, the transition is from the 5^{ th} orbit to the 2^ {nd} orbit. It belongs to the Balmer series. Wave number \bar{v} for the transition is given by,\\ 1.097*10^7m^{-1}(\dfrac{1}{2^2}-\dfrac{1}{5^2})\\ =1.097*10^7m^{-1}(\dfrac{21}{100})\\ =2.303*106m^{-1} \\ Wavelength (\lambda ) associated with the emission transition is given by, \lambda =\dfrac{1}{v}\\ =\dfrac{1}{2.303*10^6m^{-1}}\\ =0.434*10^{-6}m\\ \lambda=434nm 57 Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6×10^ 6 ms^{ –1}, calculate de Broglie wavelength associated with this electron. ##### Solution : From de Broglie’s equation,\\ \lambda=\dfrac{h}{mv}\\ \lambda=\dfrac{6.626*106{-34}Js}{(9.10939*10^{-31}kg)(1.6*10^6ms^{-1})}\\ =4.55*10^{-10}m\\ \lambda=455 pm \\ de Broglie’s wavelength associated with the electron is 455 pm. 58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron. ##### Solution : From de Broglie’s equation,\\ \lambda=\dfrac{h}{mv}\\ v=\dfrac{h}{m \lambda} \\ Where,\\ v = velocity of particle (neutron)\\ h = Planck’s constant\\ m = mass of particle (neutron)\\ \lambda = wavelength\\ Substituting the values in the expression of velocity (v),\\ \lambda=\dfrac{(6.626*10^{-34})Kgm^2s^{-1}}{(1.675*10^{-27}kg)(8*10^{-10}m)}\\ =\dfrac{6.626*10^3}{1.675*8 }ms^{-1}\\ =4.94*10^2 ms^{-1}\\ v=494 ms^{-1} \\ Velocity associated with the neutron = 494 ms^{ –1} 59 If the velocity of the electron in Bohr’s first orbit is 2.19×10 ^6 ms ^{–1}, calculate the de Broglie wavelength associated with it. ##### Solution : According to de Broglie’s equation,\\ \lambda=\dfrac{h}{mv} \\ Where,\\ \lambda =wavelength associated with the electron\\ h = Planck’s constant\\ m = mass of electron\\ v = velocity of electron\\ Substituting the values in the expression of \lambda:\\ \lambda=\dfrac{6.626810^{-34}Js}{(9.10939*10^{-31}kg)(2.19*10^6 ms^{-1})}\\ \lambda=3.32*10^{-10}m=3.32*10^{-10}m*\dfrac{100}{100}\\ \lambda=332*10^{-12}m\\ \lambda=332pm \\ Wavelength associated with the electron = 332 pm 60 The velocity associated with a proton moving in a potential difference of 1000 V is 4.37×10 ^5 ms^{ –1}. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity. ##### Solution : According to de Broglie’s expression,\\ \lambda=\dfrac{h}{mv} \\ Substituting the values in the expression,\\ \lambda=\dfrac{6.626*10^{-34}Js}{(0.1kg)(4.37*10^5ms^{-1})}\\ \lambda=1.516*10^{-38}m \\ 61 If the position of the electron is measured within an accuracy of +0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4\pi m × 0.05 nm, is there any problem in defining this value. ##### Solution : From Heisenberg’s uncertainty principle,\\ \Delta x*\Delta p=\dfrac{h}{4 \pi}=\Delta p=\dfrac{1}{\Delta x}.\dfrac{h}{4 \pi} \\ Where,\\ \Delta x = uncertainty in position of the electron\\ \Delta p = uncertainty in momentum of the electron\\ Substituting the values in the expression of \Delta p:\\ \Delta p=\dfrac{1}{0.002nm}*\dfrac{6.626*10^{-34}Js}{4*(3.14)}\\ =\dfrac{1}{2*10^{-12}m}*\dfrac{6.626*10^{-34}Js}{4*(3.14)}\\ =2.637*10^{23}Js m^{-1}\\ \Delta p=2.637*10^{-23}kg ms^{-1}(1 J=1 kg ms^2s^{-1}) Uncertainty in the momentum of the electron = 2.637×10 ^{–23} kg ms ^{–1}.\\ Actual momentum =\dfrac{h}{4 \pi_m* 0.05nm}\\ =\dfrac{6.626*10^{-34}Js}{4*3.14*5.0*10^{-11}m}\\ =1.055*10^{-24}kg ms^{-1}$$\\$ Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

62   The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?

##### Solution :

Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases. Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atom with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.

63   Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?$\\$ (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p

##### Solution :

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi- electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron(s) in it.$\\$ (i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.$\\$ (ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.$\\$ (iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.$\\$

64   The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

##### Solution :

Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom. The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

65   Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

##### Solution :

(a) Phosphorus (P): Atomic number = 15 The electronic configuration of P is: $1s ^2 2s ^2 2p ^6 3s ^2 3p ^3$ The orbital picture of P can be represented as:

From the orbital picture, phosphorus has three unpaired electrons. (b) Silicon (Si): Atomic number = 14 The electronic configuration of Si is: $1s^ 2 2s ^2 2p ^6 3s^ 2 3p ^2$$\\The orbital picture of Si can be represented as: From the orbital picture, silicon has two unpaired electrons.\\ (c) Chromium (Cr): Atomic number = 24\\ The electronic configuration of Cr is: 1s ^2 2s^ 2 2p^ 6 3s ^2 3p^ 6 4s^ 1 3d ^5$$\\$ The orbital picture of chromium is:

From the orbital picture, chromium has six unpaired electrons.$\\$ (d) Iron (Fe): Atomic number = 26$\\$ The electronic configuration is: $1s ^2 2s^ 2 2p ^6 3s^ 2 3p^ 6 4s ^2 3d ^6$$\\ The orbital picture of chromium is:\\ From the orbital picture, iron has four unpaired electrons.\\ (e) Krypton (Kr): Atomic number = 36\\ The electronic configuration is: 1s ^2 2s ^2 2p ^6 3s^ 2 3p ^6 4s^ 2 3d ^{10} 4p ^6$$\\$ The orbital picture of krypton is:$\\$

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

66   (a) How many sub-shells are associated with n = 4?$\\$ (b) How many electrons will be present in the sub-shells having ms value of $–1/2$ for $n = 4?$$\\$

##### Solution :

(a) n = 4 (Given)$\\$ For a given value of ‘n’, ‘l’ can have values from zero to (n – 1). $\therefore$ l = 0, 1, 2, 3$\\$ Thus, four sub-shells are associated with n = 4, which are s, p, d and f.$\\$ (b) Number of orbitals in the $n^{ th}$ shell = $n ^2$ For n = 4$\\$Number of orbitals = 16$\\$ If each orbital is taken fully, then it will have 1 electron with ms value of$(-\dfrac{1}{2})$.$\\$ $\therefore$Number of electrons with ms value of$(-\dfrac{1}{2})$is 16.