Modern Physics Ques 104
- A metal plate of area $1 \times 10^{-4} m^{2}$ is illuminated by a radiation of intensity $16 $ $m $ $ W / m^{2}$. The work function of the metal is $5 $ $eV$. The energy of the incident photons is $10 $ $eV$ and only $10 \%$ of it produces photoelectrons. The number of emitted photoelectrons per second and their maximum energy, respectively will be (Take, $\left.1 eV=1.6 \times 10^{-19} J\right)$
(a) $10^{11}$ and $5 $ $eV$
(b) $10^{12}$ and $5 $ $eV$
(c) $10^{10}$ and $5\ \text{eV}$
(d) $10^{14}$ and $10\ eV$
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Answer:
Correct Answer: 104.(A)
Solution:
Formula:
We know that, intensity of a radiation $I$ with energy $E$ incident on a plate per second per unit area is given as
$ \Rightarrow \quad I=\frac{d E}{d A \times d t} \Rightarrow \frac{d E}{d t}=I d A \text { or } I d A $
i.e., energy incident per unit time $=I A$
Substituting the given values, we get
$ \begin{aligned} & \frac{d E}{d t}=16 \times 10^{-3} \times 1 \times 10^{-4} \\ & \frac{d E}{d t}=16 \times 10^{-7} W \quad …….(i) \end{aligned} $
Using Einstein’s photoelectric equation, we can find kinetic energy of the emitted electrons as
$ \begin{alignedat} & E=\frac{1}{2} m v^{2}+\phi \\ & \text { or } \quad(\text { Here, } \phi \text { is work function of metal} \\ &E=KE+\phi \\ & KE=E-\phi=10 eV-5 eV \\ & KE=5 eV \quad …….(ii) \end{aligned} $
Now, power per unit time for incident photons will be
$\because \quad E =N h \nu $
$ \therefore \quad \frac{d E}{d t} =h \nu \frac{d N}{d t} \text { or } h \nu \dot{N} \quad …….(iii)$
From Eqs. (i) and (iii), we get
$ \begin{aligned} & \quad h \cup \dot{N}=16 \times 10^{-7} \text { or } E \dot{N}=16 \times 10^{-7} \\ & \text { But } \quad E=10 \text{ eV} \text {, so } \\ & \dot{N}\left(1.6 \times 10^{-18}\right)=16 \times 10^{-7} \Rightarrow \dot{N}=10^{12} \end{aligned} $
$\because$ Only $10%$ of incident photons emit electrons.
So, the number of emitted electrons per second is
$ \frac{10}{100} \times 10^{12}=10^{10} $
BETA
