Modern Physics 2 Question 7
7. 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 $5 eV$
(d) $10^{14}$ and $10 eV$
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Solution:
- We know that, intensity of a radiation $I$ with energy ’ $E$ ’ incident on a plate per second per unit area is given as
$$ \Rightarrow \quad I=\frac{d E}{d A \times d t} \Rightarrow \frac{d E}{d t}=I d A \text { or } I A $$
i.e., energy incident per unit time $=I A$
Substituting the given values, we get
$$ \begin{aligned} & \frac{d E}{d t}=16 \times 10^{-3} \times 1 \times 10^{-4} \\ & \frac{d E}{d t}=16 \times 10^{-7} W \end{aligned} $$
Using Einstein’s photoelectric equation, we can find kinetic energy of the incident radiation as
$$ \begin{aligned} & E=\frac{1}{2} m v^{2}+\varphi \\ & \\ & \text { or } \quad(\text { Here, } \varphi \text { is work } \\ &=KE+\varphi \\ & KE=E-\varphi=10 eV-5 eV \\ & KE=5 eV \end{aligned} $$
Now, energy per unit time for incident photons will be
$$ \begin{array}{rlrl} \because & E & =N h \nu \\ & \therefore & \frac{d E}{d t} & =h \nu \frac{d N}{d t} \text { or } h \nu \dot{N} \end{array} $$
From Eqs. (i) and (iii), we get
$$ \begin{aligned} & \quad h \cup \dot{N}=16 \times 10^{-7} \text { or } E \dot{N}=16 \times 10^{-7} \\ & \text { But } \quad E=10 eV \text {, so } \\ & \dot{N}\left(10 \times 1.6 \times 10^{-19}\right)=16 \times 10^{-7} \Rightarrow \dot{N}=10^{12} \end{aligned} $$
$\because$ Only $10 %$ of incident photons emit electrons.
So, emitted electrons per second are
$$ \frac{10}{100} \times 10^{12}=10^{11} $$
