Question

In a photoelectric experiment, the wavelength of the incident light is decreased from $6000\stackrel{o}{A}$ to $4000\stackrel{o}{A}$. While the intensity of radiations remains the same

• A. The cut-off potential will decrease
• B. The cut-off potential will increase
• C. The photoelectric current will increase
• D. The kinetic energy of the emitted electrons will increase

Answer 1

Answer: (B), (D)

The cut-off potential will increase
The kinetic energy of the emitted electrons will increase

If energy supplied by the incident light is $E$, it is given by
$E=\frac{hc}{\lambda }$
Using the Einstein's equation,

$\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=\frac{1}{2}m{v}^2=e{V}_0$

This K.E. is equivalent to $e{V}_0$ where $V_0$ is the cut-ff potential.
From this equation we can see that as $\lambda$ decreases, $\frac{hc}{\lambda }$ increases, hence KE increases (Option D is correct).
Also, as $\lambda$ decreases, $V_0$ also increases from the equation.