Question

A certain metal when irradiated to light ($v=\,3.2\,\times \,{{10}^{16}}\,Hz$) emits photoelectrons with kinetic energy double to the of kinetic energy of photoelectrons which was emitted by the same metal when irradiated by light ($v=\,2.0\,\times \,{{10}^{16}}\,Hz$). The ${{v}_{o}}$(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: Before solving this question, we should know the formula of threshold frequency. It is represented by ${{v}_{o}}$.
${{v}_{o}}=\,\,\dfrac{W}{h}$ With the help of this formula, we can now find the threshold frequency by putting the values in the formula and solve this question.

Complete step-by-step answer: The work function is the amount of energy that is required to remove the electron from the metal’s surface. It is represented by $W$. Its formula is –
 \[W=h{{v}_{\circ }}\]
 where h is the Planck constant and \[{{v}_{\circ }}\]is the threshold frequency
The maximum Kinetic Energy can be represented by
 $\begin{align}
  & E\,=\,hv\,-\,W \\
 & \\
\end{align}$
h is the Planck constant
v is the frequency of the photon
W is the work function
Whereas, $W\,=h{{v}_{0}}$
So, $E\,=\,hv\,-\,h{{v}_{0}}$

v before the K.E becomes double - ${{v}_{1\,}}\,=\,3.2\,\times \,{{10}^{16\,}}\,Hz$
v after the K.E becomes double - ${{v}_{2\,}}=\,\,2.0\,\times \,{{10}^{16}}\,Hz$
Let’s recall the formula of kinetic energy :
$(K.E)\,=\,h(v-{{v}_{0}})$
K.E before :
${{(K.E)}_{1}}\,=\,\,h{{v}_{1}}\,-\,h{{v}_{0}}$
K.E after :
${{(K.E)}_{2\,}}=\,\,h{{v}_{2\,}}-\,h{{v}_{0}}$
As the condition is given in the question :
${{(Kinetic\,Energy)}_{1}}$ becomes twice of ${{(Kinetic\,Energy)}_{2}}$
${{(K.E)}_{1\,}}\,=\,\,2{{(K.E)}_{2}}$
$\begin{align}
  & h{{v}_{1\,}}-\,\,h{{v}_{0\,}}\,=\,\,2(\,h{{v}_{2\,}}-\,h{{v}_{0}}) \\
 & \\
\end{align}$
Multiplying 2 on the right side
$h{{v}_{0\,}}\,=\,2h{{v}_{2\,}}-\,h{{v}_{1}}$
Canceling $h$on both the sides
${{v}_{0\,}}\,=\,2{{v}_{2}}\,-\,{{v}_{1}}$
${{v}_{0}}$= $2\,\times $($2\,\times \,{{10}^{16}}$) – ($3.2\,\times \,{{10}^{16}}$)
${{v}_{0}}\,=\,8\,\times \,{{10}^{15\,}}\,Hz$

So, Option (B) ${{v}_{0}}\,=\,8\,\times \,{{10}^{15\,}}\,Hz$ is correct.

Note: After absorbing the electromagnetic radiation, the electrons leave the metal’s surface. This effect is called the photoelectric effect. There are some conditions for it : There is a need for a minimum value that needs to be obtained for the photoelectric effect to occur. The intensity of light is not responsible for the maximum kinetic energy, It depends entirely on the frequency of the light.