# Concept Of Physics Photoelectric Effect and Wave-Particle Duality

#### H C Verma

1.   Visible light has wavelengths in the range of $400\ nm$ to $780\ nm$. Calculate the range of energy of the photons of visible light.

$\lambda_1 = 400\ nm\ \ to\ \ \lambda_2 = 780\ nm$

$E = h\nu = \dfrac{hc}{\lambda} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = 6.63 \times 10^{–34}\ j - s,\ c = 3 \times 10^8 m/s,\ \lambda_1 = 400\ nm,\ \lambda_2 = 780\ nm$

$E_1 = \dfrac{6.63 \times 10}{400 \times 10^{-9}} = \dfrac{6.63 \times 3}{4} \times 10^{-19} = 5 \times 10^{-19} = 5 \times 10^{–19}\ J$ $\\$ $E_2 = \dfrac{6.63 \times 3}{7.8} \times 10^{-19} = 2.55 \times 10^{-19}\ J$ $\\$ So, the range is $5 \times 10^{–19}\ J$ to $2.55 \times 10^{–19}\ J$

2.   Calculate the momentum of a photon of light of wavelength $500\ nm$.

$\lambda = \dfrac{h}{p}$ $\\$ $\Rightarrow P = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10 ^{-34}}{500 \times10^{-9}} J - S = 1.326 \times 10^{-27} = 1.33 \times 10^{–27}\ kg – m/s.$

3.   An atom absorbs a photon of wavelength $500\ nm$ and emits another photon of wavelength $700\ nm$. Find the net energy absorbed by the atom in the process

$\lambda_1 = 500\ nm = 500 \times 10^{–9}\ m, \lambda_2 = 700\ nm = 700 \times 10{–9}\ m$ $\\$ $E_1 – E_2 =$ Energy absorbed by the atom in the process. $= hc\ \Big[\dfrac{1}{\lambda_1} – \dfrac{1}{\lambda_2}\Big]$ $\\$ $\Rightarrow 6.63 \times 3\Big[\dfrac{1}{5} – \dfrac{1}{7}\Big] \times 10^{–19} = 1.136 \times 10^{–19}\ J$

4.   Calculate the number of photons emitted per second by a $10\ W$ sodium vapour lamp. Assume that $60$% of the consumed energy is converted into light. Wavelength of sodium light $= 590\ nm$.

$P = 10\ W \ \ \ \ \therefore E\ in\ 1\ sec = 10\ J \ \ \ \ \ \ \ \ \ \ \$ % used to convert into photon $= 60$% $\\$ $\therefore$ Energy used $= 6\ J$ $\\$ Energy used to take out $1$ photon $= \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{590 \times 10^{-9}} = \dfrac{6.633}{590} \times 10^{-17}$ $\\$ No. of photons used $= \dfrac{6}{\dfrac{6.63 \times 3}{590}\times 10^{-17}} = \dfrac{6 \times 590}{6.63 \times 3} \times 10^{17} = 176.9 \times 10^{17} = 1.77 \times 10^{19}$

5.   When the sun is directly overhead, the surface of the earth receives $1\cdot4 \times 10^3\ W/m^2$ of sunlight. Assume that the light is monochromatic with average wavelength $500\ nm$ and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is $1\cdot5\ \times 10^{11}\ m$. (a) Calculate the number of photons falling per second on each square metre of earth's surface directly below the sun. (b) How many photons are there in each cubic metre near the earth's surface at any instant ? (c) How many photons does the sun emit per second ?

a) Here intensity $= I = 1.4 \times 10^3 \omega/m^2$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ Intensity, $I = \dfrac{power}{area} = 1.4 \times 10^{3} \omega/m^2$ $\\$ Let no.of photons/sec emitted $= n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ $\therefore$ Power = Energy emitted/sec $= \dfrac{nhc}{\lambda} = P$ $\\$ No.of $photons/m^2 = \dfrac{nhc}{\lambda} =$ intensity $\\$ $N = \dfrac{}{} = \dfrac{1.9 \times 10^3 \times 5 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8} = 3.5 \times 10^{21}$ $\\$ b) Consider no.of two parts at a distance $r$ and $r + dr$ from the source. $\\$ The time interval $‘dt’$ in which the photon travel from one point to another $= \dfrac{dv}{e} = dt$. $\\$ In this time the total no.of photons emitted $= N = n\ dt = \Big(\dfrac{p\lambda}{hc}\Big)\dfrac{dr}{C}$ $\\$ These points will be present between two spherical shells of radii $‘r$’ and $r + dr$. It is the distance of the $1^st$ point from the sources. No.of photons per volume in the shell. $\\$

$(r + r + dr) = \dfrac{N}{2\pi{r}2dr} = \dfrac{P\lambda dr}{hc^2} = \dfrac{1}{4\pi{r^2}ch} = \dfrac{p\lambda}{4\pi{h}c^2r^2}$ $\\$ In the case $= 1.5 \times 10^{11}\ m,\ \lambda = 500\ nm,\ = 500 \times 10^{–9}\ m$ $\\$ $\dfrac{P}{4\pi{r^2}} = 1.4 \times 10^{3}\ ,\ \therefore No.of\ photons/m^3\ = \dfrac{P}{4\pi{r^2}} \dfrac{\lambda}{hc^2}$ $\\$ $= 1.4 \times 10^3 \times \dfrac{500 \times10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8 } = 1.2 \times 10^{13}$ $\\$ No.of photons $= (No.of\ photons/sec/m^2) \times$ Area $\\$ $= (3.5 \times 10^{21}) \times 4\pi{r^2}$ $\\$ $= 3.5 \times 10^{21} \times 4(3.14)(1.5 \times 10^{11})^2 = 9.9 \times 10^{44}$

6.   A parallel beam of monochromatic light of wavelength $663\ nm$ is incident on a totally reflecting plane mirror. The angle of incidence is $60°$ and the number of photons striking the mirror per second is $1\cdot0 \times 10^{19}$ . Calculate the force exerted by the light beam on the mirror.

$\lambda = 663 \times 10^{–9}\ m,\ \theta = 60°,\ n = 1 \times 10^{19}, \lambda = \dfrac{h}{p}$ $\\$ $\Rightarrow P = \dfrac{p}{\lambda} = 10^{–27}$ $\\$ Force exerted on the wall $= n(mv\ cos\ \theta –(–mv\ cos\ \theta)) = 2n\ mv\ cos\theta.$ $\\$ $= 2 \times 1 \times 10^{19} \times 10^{–27} \times \dfrac{1}{2} = 1 \times 10^{–8}\ N$.

7.   A beam of white light is incident normally on a plane surface absorbing $70$% of the light and reflecting the rest. If the incident beam carries $10\ W$ of power, find the force exerted by it on the surface.

Power $= 10\ W \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ P $\rightarrow$ Momentum

$\lambda = \dfrac{h}{p} \ \ \ \ \ \ \ \ \ \ \ \ or, P = \dfrac{hc}{\lambda} =$ Power $(W) \ \ \ \ \ \ \ \ \ \ \ \ \ \ or, \dfrac{P}{t} = \dfrac{h}{\lambda{t}}$

$E =\dfrac{hc}{\lambda} \ \ \ \ \ \ \ \ \ \ \ \ \ or, \dfrac{E}{t} = \dfrac{hc}{\lambda{t}} =$ Power $(W)$ $\\$ $W = \dfrac{Pc}{t} \ \ \ \ \ \ \ \ \ or, \dfrac{P}{t} = \dfrac{W}{c} =$ force. $\\$ or Force $\ \ \ \ \ \ \ \ \ = \dfrac{7}{10}\ (absorbed) + 2 \times \dfrac{3}{10}$ (reflected) $\\$ $= \dfrac{7}{10} \times \dfrac{W}{C} + 2 \times \dfrac{3}{10} \times \dfrac{W}{C} \Rightarrow \dfrac{7}{10} \times \dfrac{10}{3 \times 10^8} + 2 \times \dfrac{3}{10} \times \dfrac{10}{3 \times 10^8}$ $\\$ $= \dfrac{13}{3} \times 10^{–8} = 4.33 \times 10^{–8}\ N$.

8.   A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light as shown in figure (42-E1). The mass of the mirror is $20\ g$. Assume that there is no absorption in the lens and that $30$% of the light emitted by the source goes through the lens. Find the power of the source needed to support the weight of the mirror. Take $g = 10\ m/s^2$.

$m = 20\ g$ $\\$ The weight of the mirror is balanced. Thus force exerted by the photons is equal to weight $\\$ $P = \dfrac{h}{\lambda} \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \dfrac{hc}{\lambda} = PC$ $\\$ $\Rightarrow \dfrac{E}{t} = \dfrac{P}{t}C$ $\\$ $\Rightarrow$$m = 20\ g \\ The weight of the mirror is balanced. Thus force exerted by the photons is equal to weight \\ P = \dfrac{h}{\lambda} \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \dfrac{hc}{\lambda} = PC \\ \Rightarrow \dfrac{E}{t} = \dfrac{P}{t}C \\ \Rightarrow Rate of change of momentum = \dfrac{Power}{C} 30% of light passes through the lens. Thus it exerts force. 70% is reflected. \\ \therefore Force exerted = 2(rate of change of momentum) \\ = 2 \times \dfrac{power}{C} \\ 30 % \Big(\dfrac{2 \times Power}{C}\Big) = mg \\ \Rightarrow = \dfrac{20 \times 10^{-3} \times 3 \times 10^8 \times 10}{2 \times 3} = 10 w = 100 MW. 9. A 100\ W light bulb is placed at the centre of a spherical chamber of radius 20\ cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. Find the pressure exerted by the light on the surface of the chamber. Power = 100\ W \\ Radius = 20\ cm \\ 60% is converted to light = 60 w \\ Now, Force = \dfrac{Power}{velocity} = 2 \times 10^{-7}\ N \\ Pressure =\dfrac{force}{area} = \dfrac{2 \times 10^{-7}}{4 \times 3.14 \times (0.2)^2} = \dfrac{1}{8 \times 3.14} \times 10^{-5} \\ = 0.039 \times 10^{–5} = 3.9 \times 10^{–7} = 4 \times 10^{–7}\ N/m^2. 10. A sphere of radius 100\ cm is placed in the path of a parallel beam of light of large aperture. The intensity of the light is 0\cdot50\ W/cm^2. If the sphere completely absorbs the radiation falling on it, find the force exerted by the light beam on the sphere. We know, If a perfectly reflecting solid sphere of radius ‘r’ is kept in the path of a parallel beam of light of large aperture if intensity is I, \\ Force = \dfrac{\pi{r^2}I}{C} \\ I = 0.5 W/m^2,\ r\ = 1\ cm,\ C = 3 \times 10^8\ m/s \\ Force = \dfrac{\pi \times (1)^2 \times 0.5}{3 \times 10^8} = \dfrac{3.14 \times 0.5}{3 \times 10^8} \\ = 0.523 \times 10^{–8} = 5.2 \times 10^{–9}\ N 11. Show that it is not possible for a photon to be completely absorbed by a free electron. For a perfectly reflecting solid sphere of radius ‘r’ kept in the path of a parallel beam of light of large aperture with intensity ‘I’, force exerted = \dfrac{\pi{r^2}I}{C} 12. Two neutral particles are kept 1\ m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible. If the i undergoes an elastic collision with a photon. Then applying energy conservation to this collision \\ We get, \dfrac{hC}{\lambda} + m_0c^2 = mc^2 \\ and applying conservation of momentum {h}{\lambda} = mv \\ Mass of e = m = \dfrac{m_0}{\sqrt{1 - v^2/ c^2}} \\ from above equation it can be easily shown that \\ V = C \ \ \ \ \ \ \ or \ \ \ \ \ \ \ V = 0 \\ both of these results have no physical meaning hence it is not possible for a photon to be completely absorbed by a free electron. 13. Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350\ nm is incident on a cesium surface. Work function of cesium = 1\cdot9\ eV. r = 1\ m Energy = \dfrac{kq^2}{R} = \dfrac{kq^2}{1} \\ Now, \dfrac{kq^2}{1} = \dfrac{hc}{\lambda} \ \ \ \ \ \ \ \ \ \ \ or, \lambda = \dfrac{hc}{kq^2} \\ For max ‘\lambda’,\ ‘q’ should be min, \\ For minimum ‘e’ = 1.6 \times 10^{–19}\ C \\ Max \lambda = \dfrac{hc}{kq^2} = 0.863 \times 10^3 = 863\ m. \\ For next smaller wavelength =\dfrac{6.63 \times 3 \times 10^{-34} \times 10^8}{9 \times 10^{9} \times (1.6 \times 2)^2 \times 10^{-38}} = \dfrac{863}{4} = 215.74\ m 14. The work function of a metal is 2.5 \times 10^{-19}\ J. (a) Find the threshold frequency for photoelectric emission, (b) If the metal is exposed to a light beam of frequency 6.0 \times 10^{14}\ Hz, what will be the stopping potential ? \lambda = 350\ nn = 350 \times 10{–9}\ m \\ \phi = 1.9\ eV \\ Max KE of electrons = \dfrac{hC}{\lambda} - \phi = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{350 \times 1.6 \times 10^{-19}} - 19 \\ = 1.65\ eV = 1.6\ eV. 15. The work function of a metal is 2\cdot5 \times 10^{-19} J. (a) Find the threshold frequency for photoelectric emission, (b) If the metal is exposed to a light beam of frequency 6\cdot0 \times 10\ Hz, what will be the stopping potential ? W_0 = 2.5 \times 10^{–19}\ J \\ a) We know W_0 = h\nu_0 \\ \nu_0 = \dfrac{W_0}{h} = \dfrac{2.5 \times 10^{19}}{6.63 \times 10^{-34}} = 3.77 \times 10^{14}\ Hz = 3.8 \times 10^{14}\ Hz \\ b) eV_0 = h\nu – W_0 \\ or, V_0 = \dfrac{}{} =\dfrac{6.63 \times 10^{-34} \times 6 \times 10^{14} - 2.5 \times 10^{-19}}{1.6 \times 10^{-19}} = 0.91\ V 16. The work function of a photoelectric material is 4\cdot0\ eV. (a) What is the threshold wavelength ? (b) Find the wavelength of light for which the stopping potential is 2\cdot5\ V. \phi = 4\ eV = 4 \times 1.6 \times 10^{–19}\ J \\ a) Threshold wavelength = \lambda \phi = \dfrac{hc}{\lambda} \\ \Rightarrow \lambda = \dfrac{hC}{\phi} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 1.6 \times 10^{-19}} = \dfrac{6.63 \times 3}{6.4} \times \dfrac{10^{-27}}{10^{-9}} = 3.1 \times 10^{-17}\ nm. \\ b) Stopping potential is 2.5\ V \\ \Rightarrow \dfrac{hc}{\lambda} = 4 \times 1.6 \times 10{–19} + 1.6 \times 10{–19} \times 2.5 \Rightarrow \lambda = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda \times 1.6 \times 10^{-19}} = 4 + 2.5 \\ \Rightarrow \dfrac{6.63 \times 3 \times 10^{-26}}{1.6 \times 10^{-19} \times 6.5} = 1.9125 \times 10^{–7} = 190\ nm. 17. Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength 400\ nm falls on a metal having work function 2\cdot5\ eV. Energy of photoelectron \\ \Rightarrow \dfrac{1}{2} mv^2 = \dfrac{hc}{\lambda} - hv_0 = \dfrac{4.14 \times 10^{-15} \times 3 \times 10^8}{4 \times 10^{-7}} = 2.5ev = 0.605\ ev \\ We know KE = \dfrac{P^2}{2m} \Rightarrow = 2m \times KE \\ P^2 = 2 \times 9.1 \times 10^{–31} \times 0.605 \times 1.6 \times 10^{–19} \\ P = 4.197 \times 10^{–25}\ kg – m/s 18. When a metal plate is exposed to a monochromatic beam of light of wavelength 400\ nm, a negative potential of 1\cdot1\ V is needed to stop the photocurrent. Find the threshold wavelength for the meta \lambda = 400\ nm = 400 \times 10{–9}\ m \\ V_0 = 1.1\ V \\ \dfrac{hc}{\lambda} = \dfrac{hc}{\lambda_0} + ev_0 \\ \Rightarrow \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-19}} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda_0} + 1.6 \times 10^{-19} \times 1.1 \Rightarrow 4.97 = \dfrac{19.98 \times 10^{-26}}{\lambda_0} + 1.76 \\ \Rightarrow \dfrac{19.98 \times 10^{-26}}{\lambda_0} = 4.97 - 17.6 = 3.21 \\ \Rightarrow \dfrac{19.98 \times 10^{-26}}{3.21} = 6.196 \times 10^{–7}\ m = 620\ nm. 19. In an experiment on photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. The data collected are as follows: \\ wavelength (nm): \ \ \ \ \ \ \ \ \ \ \ \ \ 350\ \ 400\ \ 450\ \ 500\ \ 550 \\ stopping potential (V): \ \ \ \ \ \ 1\cdot45\ \ 1\cdot00\ \ 0\cdot66\ \ 0\cdot38\ \ 0\cdot16 \\ Plot the stopping potential against inverse of wavelength (1/\lambda) on a graph paper and find (a) the Planck constant, (b) the work function of the emitter and (c) the threshold wavelength. a) When \lambda = 350,\ V_s = 1.45 \\ and when \lambda = 400,\ V_s = 1 \\ \therefore \dfrac{hc}{350} = W + 1.45 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .......(1) \\ and \dfrac{hc}{400} = W + 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .......(2) \\ Subtracting (2) from (1) and solving to get the value of h we get h = 4.2 \times 10^{–15}\ ev-sec \\ b) Now work function = w = \dfrac{hc}{\lambda} = ev - s \\ \dfrac{1240}{350} - 1.45 = 2.15 ev. \\ c) w = \dfrac{hc}{\lambda} = \lambda_{there cathod} = \dfrac{hc}{w} \\ = \dfrac{}{} = 576.8\ nm 20. The electric field associated with a monochromatic beam becomes zero 1\cdot2 \times 10 'J times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2\cdot0\ eV. The electric field becomes 0\ 1.2 \times 10^{45} times per second. \\ \therefore Frequency = \dfrac{}{} = 0.6 \times 10^{15} \\ h\nu = \phi_0 + kE \\ \Rightarrow h\nu - \phi_0 = KE \\ \Rightarrow KE = \dfrac{6.63 \times 10^{-34} \times 0.6 \times 10^{15}}{1.6 \times 10^{-19}} - 2 \\ = 0.482\ ev = 0.48\ ev. 21. The electric field associated with a light wave is given by$$E = E_0\ sin\ [(1\cdot57 \times 10^7 ; m ') (x - ct)].$$Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1\cdot9\ eV. E = E_0\ sin\ [(1.57 \times 10^7\ m^{–1}) (x\ –\ ct)] \\ W = 1.57 \times 10^{7} \times C \Rightarrow f = \dfrac{1.57 \times 10^7 \times 3 \times 10^8}{2\pi}Hz \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ W_0 = 1.9\ ev \\ Now eV_0 = h\nu – W_0 \\ = 4.14 \times 10^{–15} \times \dfrac{1.57 \times 3 \times 10^{15}}{2\pi} – 1.9\ ev \\ = 3.105 – 1.9 = 1.205\ ev \\ So, V_0 =\dfrac{1.205 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.205\ V. 22. The electric field at a point associated with a light wave is E - (100\ V/m)\ sin\ [(3\cdot0 \times 10^{15} s^{-1})t]\ sin\ [(6\cdot0 \times 10^{15} s^{-1})t]. If this light falls on a metal surface having a work function of 2\cdot0\ eV, what will be the maximum kinetic energy of the photoelectrons ? E = 100\ sin\ [(3 \times 10^{15}\ s^{–1})t]\ sin\ [6 \times 10^{15}\ s^{–1})t] \\ = 100\ \dfrac{1}{2} [cos[(9 \times 10^{15}\ s^{–1})t] – cos [3 \times 10^{15}\ s^{–1})t] \\ The w are 9 \times 10^{15} and 3 \times 10^{15} \\ fotr largest K.E. \\ f_{max} = \dfrac{w_{max}}{2\pi} = \dfrac{9 \times 10^{15}}{2\pi} \\ E – \phi_0 = K.E. \\ \Rightarrow hf – \phi_0 = K.E. \\ \Rightarrow \dfrac{6.63 \times 10^{-34} 9 \times 10^{15}}{2\pi \times 1.6 \times 10^{-19}} = 2 - KE \\ \Rightarrow KE = 3.938\ ev = 3.93\ ev. 23. A monochromatic light source of intensity 5\ mW emits 8 \times 10^{15} photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2\cdot0\ V. Calculate the work function of the metal. W_0 = hv – ev_0 \\ = \dfrac{5 \times 10^{-3}}{8 \times 10^{15}} - 1.6 \times 10^{-19} \times 2 \ \ \ \ \ \ \ \ (Given V_0 = 2V, No. of photons = 8 \times 10^{15}, Power = 5\ mW) \\ = 6.25 \times 10^{–19} – 3.2 \times 10^{–19} = 3.05 \times 10^{–19}\ J \\ = \dfrac{3.05 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.906\ eV. 24. Figure (42-E2) is the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function. We have to take two cases : \\ Case I … \ \ \ v_0 = 1.656 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nu = 5 \times 10^{14}\ Hz \\ Case II… \ \ \ v_0 = 0 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nu = 1 \times 10^{14}\ Hz We know ; \\ a) ev_0 = h\nu - w_0 \\ \ \ \ 1.656e = h \times 5 \times 10^{14}\ –\ w_0 \ \ \ \ \ \ \ \ \ \ \ \ …(1) \\ \ \ \ 0 = 5h \times 10^{14}\ –\ 5w_0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ …(2) \\ \ \ \ 1.656e = 4w_0 \Rightarrow w_0 = \dfrac{1.656}{4}\ ev = 0.414\ ev \\ b) Putting value of w_0 in equation (2) \\ \Rightarrow 5w_0 = 5h \times 10^{14} \\ \Rightarrow 5 \times 0.414 = 5 \times h \times 10^{14} \\ \Rightarrow h = 4.414 \times 10^{-15}\ ev-s 25. A photographic film is coated with a silver bromide layer. When light falls on this film, silver bromide molecules dissociate and the film records the light there. A minimum of 0\cdot6\ eV is needed to dissociate a silver bromide molecule. Find the maximum wavelength of light that can be recorded by the film. w_0 = 0.6\ ev For w_0 to be min ‘\lambda’ becomes maximum.\\ w_0 = \dfrac{hc}{\lambda} or \lambda = \dfrac{hc}{w_0} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.6 \times 1.6 \times 1)^{-19}} \\ = 20.71 \times 10^{–7}\ m\ = 2071\ nm 26. In an experiment on photoelectric effect, light of wavelength 400\ nm is incident on a cesium plate at the rate of 5\cdot0\ W. The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches its saturation value. Assuming that on the average one out of every 10 photons is able to eject a photoelectron, find the photocurrent in the circuit. \lambda = 400\ nm,\ P = 5\ w \\ E of 1 photon = \dfrac{}{} = \Big(\dfrac{1242}{400}\Big)\ ev \\ No.of electrons =\dfrac{5}{Energy\ of\ 1\ photon} = \dfrac{5 \times 400}{1.6 \times 10^{-18} \times 1242} \\ No.of electrons = 1 per 10^6 photon. \\ No.of photoelectrons emitted = \dfrac{5 \times 400}{1.6 \times 1242 \times 10^{-19} \times 10^6} \\ Photo electric current = \dfrac{5 \times 400}{1.6 \times 10^{-18} \times 1242} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-6} A = 1.6 \mu{A} \\ 27. A silver ball of radius 4\cdot8\ cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200\ nm is incident on the ball for some time during which a total light energy of 1\cdot0 \times 10^{-}\ J falls on the surface. Assuming that on the average one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball assuming zero potential at infinity. What is the potential at the centre of the ball ? \lambda = 200\ nm = 2 \times 10^{–7}\ m \\ E of one photon = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7}} = 9.945 \times 10^{–19} \\ No.of photons = \dfrac{1 \times 10^{-17}}{9.945 \times 10^{-19}} = 1 \times 10^{11}\ no.s \\ Hence, No.of photo electrons = \dfrac{1 \times 10^{11}}{10^4} = 1 \times 10^7 \\ Net amount of positive charge ‘q’ developed due to the outgoing electrons \\ = 1 \times 10^7 \times 1.6 \times 10^{–19} = 1.6 \times 10^{–12}\ C. \\ Now potential developed at the centre as well as at the surface due to these charger \\ = \dfrac{Kq}{r} = \dfrac{9 \times 10^9 \times 1.6 \times 10^{-12}}{4.8 \times 10^{2}} = 3 \times 10^{–1}\ V = 0.3\ V . 28. In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of 10\ cm and are connected through an ammeter without any cell (figure 42-E3). A magnetic field B exists parallel to the plates. The work function of the emitter is 2\cdot39\ eV and the light incident on it has wavelengths between 400\ nm and 600\ nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge. \lambda_0 = 2.39\ eV \\ \lambda_1 = 400\ nm,\ \lambda_2 = 600\ nm \\ for B to the minimum energy should be maximum \\ \therefore \lambda should be minimum. \\ E = \dfrac{hc}{\lambda}\phi_0 = 3.105 – 2.39 = 0.715\ eV. \\ The presence of magnetic field will bend the beam there will be no current if the electron does not reach the other plates. \\ \ \ \ \ r = \dfrac{mv}{qB} \\ \Rightarrow r = \dfrac{\sqrt{2mE}}{qB} \\ \Rightarrow 0.1 = \dfrac{\sqrt{2 \times 9.1 10^{-31} \times1.6 \times 10^{-19} \times 0.715}}{1.6 \times 10^{-19} \times B} \\ \Rightarrow B = 2.85 \times 10^{–5}\ T 29. In the arrangement shown in figure (42-E4), y = 1\cdot0\ mm,\ d =0\cdot24\ mm and D = 1\cdot2\ m. The work function of the material of the emitter is 2\cdot2\ eV. Find the stopping potential V needed to stop the photocurrent. Given : fringe width, \\ y = 1.0\ mm \times 2 = 2.0 mm,\ D = 0.24\ mm,\ W_0 = 2.2 ev,\ D = 1.2\ m \\ y = \dfrac{\lambda{D}}{d} \\ or, \lambda = \dfrac{yd}{D} = \dfrac{2 \times 10^{-3} \times 0.24 \times 10^{-3}}{1.2} = 4 \times 10^{–7}\ m. \\ E = \dfrac{hc}{\lambda} = \dfrac{4.14 \times 10^{-15} \times 3 \times 10^8}{4 \times 10} = 3.105\ ev \\ Stopping potential eV-0 = 3.105 – 2.2 = 0.905\ V 30. In a photoelectric experiment, the collector plate is at 2\cdot0\ V with respect to the emitter plate made of copper (\psi = 4\cdot5\ eV). The emitter is illuminated by a source of monochromatic light of wavelength 200\ nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector. \phi = 4.5\ eV,\ \lambda = 200\ nm \\ Stopping potential or energy = E – \phi = \dfrac{WC}{\lambda} - \phi \\ Minimum 1.7\ V is necessary to stop the electron \\ The minimum K.E. = 2eV \\ [Since the electric potential of 2 V is reqd. to accelerate the electron to reach the plates] \\ the maximum K.E. = (2+1, 7)ev = 3.7\ ev. 31. A small piece of cesium metal (\psi = 1\cdot9\ eV) is kept at a distance of 20\ cm from a large metal plate having a charge density of 1\cdot0 \times 10^{-9} C/m^2 on the surface facing the cesium piece. A monochromatic light of wavelength 400\ nm is incident on the cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present. Given \sigma = 1 \times 10^{–9}\ cm^{–2},\ W_0\ (Cs) = 1.9\ eV,\ d = 20 cm = 0.20 m, \lambda = 400\ nm \\ we know \rightarrow Electric potential due to a charged plate = V = E \times d \\ Where E \rightarrow elelctric field due to the charged plate = \sigma/E_0 \\ d \rightarrow Separation between the plates. \\ V = \dfrac{\sigma}{E_0} \times d = \dfrac{1 \times 10^{-9} \times 20}{8.85 \times 10^{-12} \times 100} = 22.598\ V = 22.6 \\ V_0e = h\nu - w_0 = \dfrac{hc}{\lambda} - w_0 = \dfrac{4.14 \times 10^{-15} \times 3 \times 10^8}{4 \times10^{-7}} - 19 \\ = 3.105 – 1.9 = 1.205\ ev \\ or, V_0 = 1.205\ V \\ As V_0 is much less than ‘V’ \\ Hence the minimum energy required to reach the charged plate must be = 22.6\ eV \\ For maximum KE, the V must be an accelerating one. \\ Hence max KE = V_0 + V = 1.205 + 22.6 = 23.8005\ ev 32. Consider the situation of the previous problem. Consider the fastest electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate. Here electric field of metal plate = E = \dfrac{P}{E_0} \\ \dfrac{1 \times 10^{-19}}{8.85 \times 10^{-31}} = 113\ v/m \\ accl. de = \phi = \dfrac{qE}{m} \\ = \dfrac{1.6 \times 10^{-19} \times 113}{9.1 \times 10^{-31}} = 19.87 \times 10^{12} \\ t = \dfrac{\sqrt{2y}}{a} = \dfrac{\sqrt{2 \times 20 \times 10^{-2}}}{19.87 \times 10^{-31}} = 1.41 \times 10^{–7}\ sec. \\ K.E. = \dfrac{hc}{\lambda} - w = 1.2\ eV \\ = 1.2 \times 1.6 \times 10^{–19}\ J [because in previous problem i.e. in problem 31 : KE = 1.2\ ev] \\ \therefore V = \dfrac{2KE}{m} = \dfrac{2 \times 1.2 \times 1.6 \times 10^{-19}}{4.1 \times 10^{-31}}= 0.665 \times 10^{–6} \\ \therefore Horizontal displacement = V_t \times t \\ = 0.655 \times 10^{–6} \times 1.4 \times 10^{–7} = 0.092\ m = 9.2\ cm 33. A horizontal cesium plate (\psi = 1\cdot9\ eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250\ nm and above. What should be the minimum value of \upsilon so that the vertically upward component of velocity is nonpositive for each photoelectron ? When \lambda = 250\ nm \\ Energy of photon = \dfrac{hc}{\lambda} \dfrac{1240}{250}= 4.96\ ev \\ \therefore K.E. = \dfrac{hc}{\lambda} - w = 4.96 – 1.9\ ev\ = 3.06\ ev \\ Velocity to be non positive for each photo electron \\ The minimum value of velocity of plate should be = velocity of photo electron \\ \therefore Velocity of photo electron = \sqrt{\dfrac{2KE}{m}} \\ = \sqrt{\dfrac{3.06}{9.1 \times 10^{-31}}} = \sqrt{\dfrac{3.06 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}= 1.04 \times 10^6\ m/sec 34. A small metal plate (work function \psi$$)$ is kept at a distance d from a singly ionized, fixed ion. A monochromatic light beam is incident on the metal plate and photoelectrons are emitted. Find the maximum wavelength of the light beam so that some of the photoelectrons may go round the ion along a circle.

Work function $= \phi,$ distance $= d$ $\\$ The particle will move in a circle $\\$ When the stopping potential is equal to the potential due to the singly charged ion at that point. $\\$ $\ \ \ \ \ \ \ \ eV_0 = \dfrac{hc}{\lambda} - \phi$ $\\$ $\Rightarrow V_0 = \Big(\dfrac{hc}{\lambda} - \phi\Big) \dfrac{1}{e} \Rightarrow \dfrac{ke}{2d} = \Big(\dfrac{hc}{\lambda} - \phi\Big) \dfrac{1}{e}$ $\\$ $\Rightarrow \dfrac{Ke^2}{2D} = \dfrac{hc}{\lambda} - \phi \Rightarrow \dfrac{hc}{\lambda} = \dfrac{Ke^2}{2d} + \phi = \dfrac{Ke^2 + 2d\phi}{2d}$ $\\$ $\Rightarrow \lambda = \dfrac{hc2d}{Ke^2 + 2d\phi} = \dfrac{2hcd}{\dfrac{1}{4\pi\epsilon_0 e^2} + 2d\phi} =\dfrac{8\pi\epsilon_0 hcd}{e^2 + 8\pi\epsilon_0 d\phi}$

35.   A light beam of wavelength $400\ nm$ is incident on a metal plate of work function $2\cdot2\ eV.$ (a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that $10$% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal, (b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal.

When $\lambda = 400\ nm$ $\\$ Energy of photon $= \dfrac{hc}{\lambda} = \dfrac{1240}{400}= 3.1\ eV$ $\\$ This energy given to electron $\\$ But for the first collision energy lost $= 3.1\ ev \times 10$% $= 0.31\ ev$ $\\$ for second collision energy lost $= 3.1\ ev \times 10$% $= 0.31\ ev$ $\\$ Total energy lost the two collision $= 0.31 + 0.31 = 0.62\ ev$ $\\$ $K.E.$ of photon electron when it comes out of metal $\\$ $= hc/\lambda$ – work function – Energy lost due to collision $\\$ $= 3.1\ ev\ –\ 2.2\ –\ 0.62 = 0.31\ ev$

b) For the $3^{rd}$ collision the energy lost $= 0.31\ ev$ $\\$ Which just equative the $KE$ lost in the $3^{rd}$ collision electron. It just comes out of the metal $\\$ Hence in the fourth collision electron becomes unable to come out of the metal $\\$ Hence maximum number of collision $= 4.$