# Photoelectric Effect and Wave-Particle Duality

## Concept Of Physics

### 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.

##### Solution :

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

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

$E_1 = \dfrac{6.63 \times 10}{400 \times 10^{-9}} = \dfrac{6.63 \times 3}{4} \times 10^{-19} = 5 \times 10^{-19} = 5 \times 10^{–19}\ J$ $\\$ $E_2 = \dfrac{6.63 \times 3}{7.8} \times 10^{-19} = 2.55 \times 10^{-19}\ J$ $\\$ 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 $500\ nm$.

##### Solution :

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

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

##### Solution :

$\lambda_1 = 500\ nm = 500 \times 10^{–9}\ m, \lambda_2 = 700\ nm = 700 \times 10{–9}\ m$ $\\$ $E_1 – E_2 =$ 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 $= 590\ nm$.

##### Solution :

$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 ?

##### Solution :

$(r + r + dr) = \dfrac{N}{2\pi{r}2dr} = \dfrac{P\lambda dr}{hc^2} = \dfrac{1}{4\pi{r^2}ch} = \dfrac{p\lambda}{4\pi{h}c^2r^2}$ $\\$ 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 $\\$ $= (3.5 \times 10^{21}) \times 4\pi{r^2}$ $\\$ $= 3.5 \times 10^{21} \times 4(3.14)(1.5 \times 10^{11})^2 = 9.9 \times 10^{44}$

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. $\\$

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.

##### Solution :

$\lambda = 663 \times 10^{–9}\ m,\ \theta = 60°,\ n = 1 \times 10^{19}, \lambda = \dfrac{h}{p}$ $\\$ $\Rightarrow P = \dfrac{p}{\lambda} = 10^{–27}$ $\\$ 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.

##### Solution :

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

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

$E =\dfrac{hc}{\lambda} \ \ \ \ \ \ \ \ \ \ \ \ \ or, \dfrac{E}{t} = \dfrac{hc}{\lambda{t}} =$ 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 $g = 10\ m/s^2$.

##### Solution :

$30$% of light passes through the lens. Thus it exerts force. $70$% is reflected. $\\$ $\therefore$ Force exerted $= 2$(rate of change of momentum) $\\$ $= 2 \times \dfrac{power}{C}$ $\\$ $30$ % $\Big(\dfrac{2 \times Power}{C}\Big) = mg$ $\\$ $\Rightarrow = \dfrac{20 \times 10^{-3} \times 3 \times 10^8 \times 10}{2 \times 3} = 10 w = 100 MW.$

##### Solution :

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.

##### Solution :

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 $= 4.$

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 $\\$ $= 3.1\ ev\ –\ 2.2\ –\ 0.62 = 0.31\ ev$