# 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.

(i) Mass of one electron $= 9.10939×10^{-31}$kg$\\$ Number of electrons that weigh $9.10939 × 10^{-31}$ kg=$1$$\\ Number of electrons that will weigh 1 g=(1*10^{-3}kg)\\ \dfrac{1}{9.10939*10^{-31}kg}(1*10^{-3}kg)$$\\$ $=0.1098*10^{-3+31}$$\\ =0.1098*10^{28}$$\\$ $=1.098 * 10^{27}$$\\ (ii) Mass of one electron = 9.10939 × 10^{-31}kg \\ Mass of one mole of electron =(6.022*10^{23})*(9.10939*10^{-31}kg)$$\\$ $=5.48*10^{-7}$kg$\\$ Charge on one electron $= 1.6022 × 10^{-19}$ coulomb$\\$ Charge on one mole of electron $=(1.6022*10^{-19}C)(6.022*10^{23})$$\\ =9.65*10^4 C 2 (i) Calculate the total number of electrons present in one mole of methane.\\ (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ^{14}C. (Assume that mass of a neutron = 1.675 × 10^{-27} kg).\\ (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH _3 at STP. Will the answer change if the temperature and pressure are changed? ##### Solution : (i) Number of electrons present in 1 molecule of methane (CH _4 )$$\\$ $\{1(6) + 4(1)\} = 10$$\\ Number of electrons present in 1 mole i.e., 6.023 × 10^{23}molecules of methane\\ =6.022*10^{23}*10=6.022*10^{24}$$\\$ (ii) (a) Number of atoms of$^{14}C$ in $1$ mole$= 6.023 × 10^{23}$$\\ Since 1 atom of 14 C contains (14 - 6) i.e.,\\ 8 neutrons, the number of neutrons in 14 g of 14 C is (6.023 ×10 ^{23 }) ×8. \\ Or, 14 g of 14 C contains (6.022 × 10^{23} × 8)neutrons.\\ Number of neutrons in 7 mg\\ =\dfrac{6.022*10^{23}*8*7 mg}{1400 mg}$$\\$ $=2.4092*10^{21}$$\\ (b) Mass of one neutron = 1.67493 × 10^{-27} kg$$\\$ Mass of total neutrons in $7$ g of $^{14}C$$\\ = (2.4092 × 10^{21} ) (1.67493 × 10 ^{-27}kg)$$\\$ $= 4.0352 × 10 kg$$\\ (iii) (a) 1 mole of NH _3 = \{1(14) + 3(1)\} g of NH_ 3$$\\$ $= 17$ g of $NH _3$$\\ = 6.022× 10^{23} molecules of NH _3$$\\$ Total number of protons present in $1$ molecule of $NH _3$$\\ = \{1(7) + 3(1)\}$$\\$ $= 10$$\\ Number of protons in 6.023 × 10^{23} molecules of NH _3$$\\$ $= (6.023 × 10 23 ) (10)$$\\ = 6.023 × 10 24$$\\$ $\Rightarrow 17$ g of $NH _3$ contains $(6.023 × 10^{24} )$ protons.$\\$ Number of protons in $34$ mg of $NH _3$$\\ =\dfrac{(6.022*10^{24}*34)mg}{1700 mg}$$\\$ $=1.2046*10^{22}$$\\ (b) Mass of one proton = 1.67493 × 10^{-27}kg$$\\$ Total mass of protons in $34$ mg of $NH _3$$\\ = (1.67493 × 10^{-27}kg) (1.2046 × 10^{22} )$$\\$ $= 2.0176 × 10^{-5} kg$$\\ The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed. 3 How many neutrons and protons are there in the following nuclei?\\ ^{13}_{6}C , \ ^{16}_{8}O, \ ^{24}_{12}Mg, \ ^{56}_{26}Fe, \ ^{88}_{38}Sr ##### Solution : ^{13}_{6}C:$$\\$ Atomic mass$= 13$$\\ Atomic number = Number of protons = 6$$\\$ Number of neutrons = (Atomic mass) - (Atomic number)$\\$ $= 13 - 6 = 7$$\\ ^{16}_{8}O:$$\\$ Atomic mass = $16$ Atomic number =$8$ Number of protons =$8$$\\ Number of neutrons = (Atomic mass) - (Atomic number)\\ = 16 - 8 = 8$$\\$ $^{24}_{12}Mg:$$\\ Atomic mass = 24$$\\$ Atomic number = Number of protons $= 12$$\\ Number of neutrons = (Atomic mass) -(Atomic number)\\ = 24 - 12 = 12$$\\$ $^{56}_{26}Fe:$$\\ Atomic mass=56$$\\$ Atomic number = Number of protons = $26$$\\ Number of neutrons = (Atomic mass) - (Atomic number)\\ = 56 - 26 = 30$$\\$ $^{88}_{38}Sr:$$\\ Atomic mass = 88$$\\$ Atomic number = Number of protons $= 38$$\\ Number of neutrons = (Atomic mass) - (Atomic number)\\ = 88 - 38 = 50 4 Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A)\\ (i) Z = 17, A = 35$$\\$ (ii)$Z = 92, A = 233$$\\ (iii) Z = 4, A = 9 ##### Solution : (i)^{35}_{17}CI$$\\$ (ii)$^{233}_{92}U$$\\ (iii)^{9}_{4}Be 5 Yellow light emitted from a sodium lamp has a wavelength (\lambda) of 580 nm. Calculate the frequency (v) and wave number ( v ) of the yellow light. ##### Solution : From the expression,\\ \lambda=\dfrac{c}{v}$$\\$ We,get,$\\$ $v=\dfrac{v}{\lambda}......(i)$$\\ Where,\\ v = frequency of yellow light\\ c = velocity of light in vacuum = 3 × 10 ^8 m/s$$\\$ $\lambda$= wavelength of yellow light = $580$ nm = $580 × 10^{-9}m$$\\ Substituting the values in expression (i):\\ v=\dfrac{3*10^8}{580*10^{-9}}=5.17*10^{14}S^{-1}$$\\$ Thus, frequency of yellow light emitted from the sodium lamp$\\$ $=5.17*10^{14}s^{-1}$$\\ Wave number of yellow light,\bar{v}=\dfrac{1}{\lambda}$$\\$ $=\dfrac{1}{580*10^{-9}}=1.72*10^{6}m^{-1}$

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) $=\dfrac{N_Ahc}{\lambda }$ $\\$ $=\dfrac{(6.023* 10^{23 }mol^{-1})(6.626*10^{-34}Js)(3*10^8 ms^{-1})}{242*10^{-9}m}$ $\\$ $=4.947*10^5 J mol^{-1}$ $\\$ $=494.7*10^3 J mol^{-1}$ $\\$ $=494 kJ mol^{-1}$