Modern Physics 2 Question 35
34. In a photoelectric effect set-up a point of light of power $3.2 \times 10^{-3} W$ emits monoenergetic photons of energy $5.0 eV$. The source is located at a distance of $0.8 m$ from the centre of a stationary metallic sphere of work function $3.0 eV$ and of radius $8.0 \times 10^{-3} m$. The efficiency of photoelectrons emission is one for every $10^{6}$ incident photons. Assume that the sphere is isolated and initially neutral and that photoelectrons are instantly swept away after emission.
$(1995,10 M)$
(a) Calculate the number of photoelectrons emitted per second.
(b) Find the ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest photoelectrons emitted.
(c) It is observed that the photoelectrons emission stops at a certain time $t$ after the light source is switched on why?
(d) Evaluate the time $t$.
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Answer:
Correct Answer: 34. (a) $2.55 eV$ (b) $4 \rightarrow 2$ (c) $-\frac{h}{\pi}$ (d) $0.814 m / s$
Solution:
- (a) Energy of emitted photons
$$ E _1=5.0 eV=5.0 \times 1.6 \times 10^{-19} J=8.0 \times 10^{-19} J $$
Power of the point source is $3.2 \times 10^{-3} W$ or $3.2 \times 10^{-3} J / s$.
Therefore, energy emitted per second,
$$ E _2=3.2 \times 10^{-3} J $$
Hence, number of photons emitted per second $n _1=\frac{E _2}{E _1}$
$$ \text { or } \quad \begin{aligned} n _1 & =\frac{3.2 \times 10^{-3}}{8.0 \times 10^{-19}} \\ n _1 & =4.0 \times 10^{15} \text { photon } / s \end{aligned} $$
Number of photons incident on unit area at a distance of $0.8 m$ from the source $S$ will be
$$ \begin{aligned} n _2 & =\frac{n _1}{4 \pi(0.8)^{2}}=\frac{4.0 \times 10^{15}}{4 \pi(0.64)} \\ & \approx 5 \times 10^{14} \text { photon } / s-m^{2} \end{aligned} $$
The area of metallic sphere over which photons will fall is
$$ A=\pi r^{2}=\pi\left(8 \times 10^{-3}\right)^{2} m^{2} \approx 2.01 \times 10^{-4} m^{2} $$
Therefore, number of photons incident on the sphere per second are
$$ n _3=n _2 A=\left(5.0 \times 10^{14} \times 2.01 \times 10^{-4}\right) \approx 10^{11} / s $$
But since, one photoelectron is emitted for every $10^{6}$ photons, hence number of photoelectrons emitted per second.
$$ n=\frac{n _3}{10^{6}}=\frac{10^{11}}{10^{6}}=10^{5} / s \text { or } n=10^{5} / s $$
(b) Maximum kinetic energy of photoelectrons
$K _{\text {max }}=$ Energy of incident photons - work function
$$ \begin{aligned} & =(5.0-3.0) eV=2.0 eV \\ & =2.0 \times 1.6 \times 10^{-19} J \end{aligned} $$
$K _{\max }=3.2 \times 10^{-19} J$
The de-Broglie wavelength of these photoelectrons will be
$$ \lambda _1=\frac{h}{p}=\frac{h}{\sqrt{2 K _{\max } m}} $$
Here, $h=$ Planck’s constant and $m=$ mass of electron
$$ \begin{aligned} \lambda _1 & =\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 3.2 \times 10^{-19} \times 9.1 \times 10^{-31}}} \\ & =8.68 \times 10^{-10}=8.68 \AA \end{aligned} $$
Wavelength of incident light $\lambda _2$ (in $\left.\AA\right)=\frac{12375}{E _1(\text { in eV })}$
or $\quad \lambda _2=\frac{12375}{5}=2475 \AA$
Therefore, the desired ratio is
$$ \frac{\lambda _2}{\lambda _1}=\frac{2475}{8.68}=285.1 $$
(c) As soon as electrons are emitted from the metal sphere, it gets positively charged and acquires positive potential. The positive potential gradually increases as more and more photoelectrons are emitted from its surface. Emission of photoelectrons is stopped when its potential is equal to the stopping potential required for fastest moving electrons.
(d) As discussed in part (c), emission of photoelectrons is stopped when potential on the metal sphere is equal to the stopping potential of fastest moving electrons.
Since, $\quad K _{\max }=2.0 eV$
Therefore, stopping potential $V _0=2 V$. Let $q$ be the charge required for the potential on the sphere to be equal to stopping potential or $2 V$. Then
$$ \begin{aligned} 2 & =\frac{1}{4 \pi \varepsilon _0} \cdot \frac{q}{r}=\left(9.0 \times 10^{9}\right) \frac{q}{8.0 \times 10^{-3}} \\ \therefore \quad q & =1.78 \times 10^{-12} C \end{aligned} $$
Photoelectrons emitted per second $=10^{5}$ [Part (a)] or charge emitted per second $=\left(1.6 \times 10^{-19}\right) \times 10^{5} C$
$$ =1.6 \times 10^{-14} C $$
Therefore, time required to acquire the charge $q$ will be
$$ t=\frac{q}{1.6 \times 10^{-14}} s=\frac{1.78 \times 10^{-12}}{1.6 \times 10^{-14}} s \quad \text { or } \quad t \approx 111 s $$
