Question

Consider a $20W$ bulb emitting light of wavelength $5000\stackrel{o}{A}$ and shining on a metal surface kept at a distance $2m$. Assume that the metal surface has work function of $2eV$ and that each atom on the metal surface can be treated as a circular disk of radius $1.5\stackrel{o}{A}$.

A) Estimate no. of photons emitted by the bulb per second. [Assume no other losses]

B) Will there be photoelectric emission?

C) How much time would be required by the atomic disk to receive energy equal to work function $\left(2eV\right)$?

D) How many photons would atomic disk receive within time duration $11.37s$?

E) Can you explain how photoelectric effect was observed instantaneously?

Answer 1

A) Given, power of bulb, $P=20W$

Wavelength of light, $\lambda =5000\stackrel{o}{A}$

Distance, $d=2m$

Work function, ${\varphi }_0=2eV$

Radius of the metal surface, which is treated as a circular disk,

$r=1.5\stackrel{o}{A}=1.5\times {10}^{-10}m$

Let the number of photons emitted by bulb per second,

$n^{\prime }=\frac{dN}{dt}$

Number of photons emitted by bulb per second is

$n^{\prime }=\frac{P}{hc/\lambda }=\frac{P\lambda }{hc}$

Substituting the values, we get

$n^{\prime }=\frac{20\times (5000\times {10}^{-10})}{(6.62\times {10}^{-34})\times (3\times {10}^8)}$

$\Rightarrow n^{\prime }=5\times {10}^{19}/sec$

$\text{Final Answer}:5\times {10}^{19}/sec$



B) Given, power of bulb, $P=20W$

Wavelength of light, $\lambda =5000\stackrel{o}{A}$

Distance, $d=2m$

Work function, ${\varphi }_0=2eV$

Radius of the metal surface, which is treated as a circular disk,

$r=1.5\stackrel{o}{A}=1.5\times {10}^{-10}m$

As the energy of the incident photon is given by,

$E=\frac{hc}{\lambda }$

Substituting the values, we get

$E=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{5000\times {10}^{-10}\times 1.6\times {10}^{-19}}$

$E=2.48eV$

As this energy is greater than $2eV$ (i.e., work function of metal surface), hence photoelectric emission takes place.

$\text{Final Answer}:$ Yes


C)

Given, power of bulb, $P=20W$

Wavelength of light, $\lambda =5000\stackrel{o}{A}$

Distance, $d=2m$

Work function, ${\varphi }_0=2eV$

Radius of the metal surface, which is treated as a circular disk,

$r=1.5\stackrel{o}{A}=1.5\times {10}^{-10}m$

Let $\Delta t$ be the time spent in getting the energy, ${\varphi }_0=$ (work function of metal).
Consider the figure, if $P$ is the power of source then energy received by atomic disc

$\frac{P}{4\pi d^2}\times \pi r^2\Delta t={\varphi }_0$

$\Rightarrow \Delta t=\frac{4{\varphi }_0{d}^2}{P{r}^2}$

$=\frac{4\times (2\times 1.6\times {10}^{-19})\times 2^2}{20\times (1.5\times {10}^{-10}{)}^2}\approx 11.37s$

$\text{Final Answer}:11.37s$


D) $\text{Find the number of photons emitted by bulb per second}.$

Given, power of bulb, $P=20W$

Wavelength of light, $\lambda =5000\stackrel{o}{A}$

Distance, $d=2m$

Work function, ${\varphi }_0=2eV$

Radius of the metal surface, which is treated as a circular disk,

$r=1.5\stackrel{o}{A}=1.5\times {10}^{-10}m$

Let the number of photons emitted by bulb per second,

$n^{\prime }=\frac{dN}{dt}$

Number of photons emitted by bulb per second is

$n^{\prime }=\frac{P}{hc/\lambda }=\frac{P\lambda }{hc}$

Substituting the values, we get

$n^{\prime }=\frac{20\times (5000\times {10}^{-10})}{(6.62\times {10}^{-34})\times (3\times {10}^8)}$

$\Rightarrow n^{\prime }=5\times {10}^{19}/sec$

$\text{Find the number of photons received by atomic disk}.$

Number of photons received by atomic disk in time $\Delta t$ is

$N=\frac{n^{\prime }\times \pi r^2}{4\pi d^2}\times \Delta t$

$=\frac{n^{\prime }{r}^2\Delta t}{4d^2}$

$N=\frac{(5\times {10}^{19})\times (1.5\times {10}^{-10}{)}^2\times 11.37}{4\times (2{)}^2}=0.80$

$\text{Final Answer}:0.80$


E) In photoelectric emission, there is collision between incident photon and free electron of the metal surface, which lasts for very short interval of time $(={10}^{-9}$ sec), hence, we can say that photoelectric emission is instantaneous.