Question

A certain metal when irradiated to light $\left( {v = 3.2 \times {{10}^{16}}Hz} \right)\;$emits photoelectrons with kinetic energy double to the kinetic energy of photoelectrons which was emitted by the same metal when irradiated by light $\left( {v = 2.0 \times {{10}^{16}}Hz} \right)\;$.
Calculate ${\nu _o }$​(threshold frequency) of metal is:
A.$1.2 \times {10^{14}}{{Hz}}$
B.$8 \times {10^{15}}{{Hz}}$
C.$1.2 \times {10^{16}}{{Hz}}$
D.$4 \times {10^{12}}{{Hz}}$

Answer 1

Hint : Using the formula the kinetic energy we can calculate kinetic energy for each of the given frequencies. Dividing the equation using the given condition in the formula will give us the value of threshold frequency.
Formula used: ${{K}}{{.E}} = {{h}}({\nu _1} - {\nu _o })$
Here K.E is kinetic energy, h is Planck's constant, ${\nu _1}$ is the frequency of a photon and ${\nu _o }$ is the threshold frequency.

Complete Step By Step Answer:
Let us write the equations for kinetic energy in each frequency case using the formula for kinetic energy of a photon.
When the frequency of the photon is $\;v = 3.2 \times {10^{16}}Hz$, let the kinetic energy of the emitted photon is ${{{K}}_1}$. Keeping all the things as it is we will keep the above values in the formula. We will get equation 1 as:
${{{K}}_1} = {{h}}\left( {3.2 \times {{10}^{16}}Hz - {\nu _o }} \right)$
When the frequency of the photon is $\;v = 2.0 \times {10^{16}}Hz$, let the kinetic energy of the emitted photon is ${{{K}}_2}$. In the same way we will put the known variable and we will get the equation as:
${{{K}}_2} = {{h}}\left( {2.0 \times {{10}^{16}}Hz - {\nu _o }} \right)$
According to the statement given in question ${{{K}}_1} = 2 \times {{{K}}_2}$. Substituting the values here we will get,
${{h}}\left( {3.2 \times {{10}^{16}}{{Hz}} - {\nu _o }} \right) = 2 \times {{h}}\left( {2.0 \times {{10}^{16}}{{Hz}} - {\nu _o }} \right)$
Canceling the terms we will get:
$3.2 \times {10^{16}}{{Hz}} - {\nu _o } = 4.0 \times {10^{16}}{{Hz}} - 2{\nu _o }$
We can rearrange the above equation as:
$2{\nu _o } - {\nu _o } = 4.0 \times {10^{16}}{{Hz}} - 3.2 \times {10^{16}}{{Hz}}$
We can solve the above equation as:
${\nu _o } = 1.2 \times {10^{16}}{{Hz}}$

Hence, the correct option is C.

Note :
Threshold frequency is the minimum amount of frequency of incident radiation that is required to eject electrons. If the incident light has a frequency less than the threshold frequency no ejection of electrons will take place, however high the kinetic energy of incident radiation is. It depends upon the nature of the metal on which the light is interacting.