Question

In the photoelectric experiment, the wavelength of the light incident on a metal is changed from $300nm$ to $400nm$. The decrease in stopping potential is close to: $\left( {\dfrac{{hc}}{e} = 1240nm - V} \right)$
A. $0.5V$
B. $1.0V$
C. $2.0V$
D. $1.5V$

Answer 1

Hint: Use the photoelectric equation and make two equations, first for wavelength $300nm$ and another for wavelength $400nm$. The speed in photoelectric experiment becomes equal to the speed of light.

Complete step by step answer:
When the electrons are made to escape from metals when light is above a certain frequency this frequency is known as threshold frequency and the effect occurred is called Photoelectric effect. The electrons which escape in this process are called Photoelectrons. In this experiment we can see that –
1. The maximum kinetic energy depends upon the frequency of light and not the intensity of light.
2. The frequency of light needed to reach a particular minimum value for photoelectrons to start escaping the metal.

From the above observations, we can observe that electromagnetic waves are emitted in packets of energy called photons.
Now, for this question we have to see what is photoelectric equation –
$hf = \phi + {E_k} \cdots \left( 1 \right)$
where, $h$ is the Planck’s constant which is equal to $6.63 \times {10^{ - 34}}Js$
$f$ is the frequency of incident light in Hertz
$\phi $ is the work function and is measured in Joules
${E_k}$ is maximum kinetic energy of the emitted electrons
The equation $\left( 1 \right)$ can be written as –
$hf = \phi + eV \cdots \left( 2 \right)$
We know that, $frequency = \dfrac{{speed}}{{wavelength}}$
Let the frequency be $f$ and wavelength be $\lambda $
For photoelectric effect, the speed is equal to the speed of light, $c$.
$\therefore f = \dfrac{c}{\lambda }$
Putting this in equation $\left( 2 \right)$, we get –
$\dfrac{{hc}}{\lambda } = \phi + eV$
For first wavelength –
$\dfrac{{hc}}{{{\lambda _1}}} = \phi + e{V_1} \cdots \left( 3 \right)$
For second wavelength –
$\dfrac{{hc}}{{{\lambda _2}}} = \phi + e{V_2} \cdots \left( 4 \right)$
Subtracting equation $\left( 3 \right)$ from equation $\left( 4 \right)$, we get –
$
  hc\left( {\dfrac{1}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right) = e\left( {{V_1} - {V_2}} \right) \\
   \Rightarrow \dfrac{{hc}}{e}\left( {\dfrac{{{\lambda _2} - {\lambda _1}}}{{{\lambda _1}{\lambda _2}}}} \right) = {V_1} - {V_2} \\
 $
According to the question, it is given that –
$\dfrac{{hc}}{e} = 1240nm - V$
$
  {\lambda _1} = 300nm \\
  {\lambda _2} = 400nm \\
 $
$\therefore {V_1} - {V_2} = 1240 - V\left( {\dfrac{{{\lambda _2} - {\lambda _1}}}{{{\lambda _1}{\lambda _2}}}} \right)$
Putting the values of wavelength in the above equation –
\[
   \Rightarrow {V_1} - {V_2} = 1240 - V\left( {\dfrac{{400 - 300}}{{400 \times 300}}} \right) \\
  {V_1} - {V_2} = 1240 - V\left( {\dfrac{{100}}{{400 \times 300}}} \right) \\
 \]
By further solving, we get –
\[{V_1} - {V_2} = 1V\]

Hence, the correct option is (B).

Note:If the light does not have enough frequency then, the photon has enough energy to overcome the work function. Then, no photoelectrons cannot be emitted. So, ensure that light does not have much frequency.