Question

A metal plate of area $1 \times {10^{ - 4}}{m^2}$ is illuminated by a radiation of intensity $16mW/{m^2}$. 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:
$\left[ {1eV = 1.6 \times {{10}^{ - 19}}} \right]$
A. ${10^{10}}$ and $5$ $eV$
B. ${10^{12}}$ and $5$ $eV$
C. ${10^{14}}$ and $10$ $eV$
D. ${10^{11}}$ and $5$ $eV$

Answer 1

Work function is the minimum energy required to remove the electron from a solid; the rest of the energy of the electron is converted into kinetic energy. Use the formula of kinetic energy and intensity to solve the question.
Where,
K.E is the maximum kinetic energy E is the energy of incident photons$\phi$is work function.

Complete step by step answer: It B given is the question that
$A = 1 \times {10^{ - 4}}{m^2}$is the area of the metal plate.
$\phi = 5ev$ is the work function.
$E =$ $10ev$is the energy of incident photons.
$I = 16 \times {10^{ - 3}}W/{m^2}$ is the intensity of reactions .
We have the formula $K{E_{\max }} = E - \phi$
Where,
$K{E_{\max }}$ is the maximum kinetic energy of emitted photons.
$E$ is the energy of the incident photon.
$\phi$ is the work function.
By substituting the given value in the above equation, we get.
$K{E_{\max }} = 10ev - 5ev$
$= 5ev.$
Now,
We know that,
$I = \dfrac{{nE}}{{At}}$
Where,
$I$ is the intensity of radiation.
$N$ is the number of photons.
$A$ is the area.
$T$ is time.
By substituting the given value in the above equation we get.
$16 \times {10^{ - 3}} = \left( {\dfrac{n}{t}} \right) \times \dfrac{{10 \times 1.6 \times 10}}{{{{10}^{ - 4}}}}$ $\left( {\because 1ev = 1.6 \times {{10}^{ - 19}}} \right)$
By rearranging it, we get
$\dfrac{n}{t} = \dfrac{{16 \times {{10}^{ - 3}} \times {{10}^{ - 9}}}}{{10 \times 1.6 \times {{10}^{ - 19}}}}$
$= \dfrac{{16 \times 10}}{{10 \times {{10}^{ - 19}}}}$
$\Rightarrow \dfrac{n}{t} = {10^{12}}$
Therefore, The number of photoelectrons emitted per second is ${10^{12}}$
And the maximum kinetic energy of the emitted photons is $5ev$
Therefore, from the above explanation the correct answer is option (B) ${10^{12}}$and $5ev$

Note: All the units should be in the same system of units therefore do not forget to convert $16mw/{m^2}$into $16 \times {10^{ - 3}}W/{m^2}$as all the rest of the questions are in $SI$system of units.