Question

A photosensitive surface is illuminated by radiations of wavelength $\lambda_{1}$, $\lambda_{2}\;(>\lambda_{1})$ and $\lambda_{3}$ one by one and photoemission is observed in each case. $\lambda_{1}$ and $\lambda_{2}$ lie in UV range and $\lambda_{3}$ in visible range. If $V_{1}$, $V_{2}$ and $V_{3}$ are stopping potentials in these cases respectively, then:

• A. $V_{1} = V_{2} = V_{3}$
• B. $V_{2} > V_{1} > V_{3}$
• C. $V_{1} > V_{2} > V_{3}$
• D. $V_{3} > V_{1} > V_{2}$

Hint:

eV = hc/lambda - work function. Shorter wavelength gives larger stopping potential; visible light has longer wavelength than UV.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In the photoelectric effect, a photon of energy \( hc/\lambda \) knocks out an electron. The stopping potential \( V \) measures the maximum kinetic energy of the emitted electrons.

Step 2: Key Formula or Approach:
\[ eV = \dfrac{hc}{\lambda} - \phi_0 \]
Here \( \phi_0 \) is the work function of the surface. It is the same for all three radiations, because the surface is the same. So a shorter wavelength gives a bigger \( V \).

Step 3: Arrange the wavelengths:
We are given \( \lambda_2 > \lambda_1 \). Both lie in the UV range, and \( \lambda_3 \) lies in the visible range.
Visible light has longer wavelengths than ultraviolet light (about 400-700 nm against 10-400 nm). So \( \lambda_3 > \lambda_2 \), and the order is \( \lambda_1 < \lambda_2 < \lambda_3 \).

Step 4: Arrange the stopping potentials:
Since \( V \) falls as \( \lambda \) grows, \( V_1 > V_2 > V_3 \). All three are positive because photoemission occurs in each case.

Step 5: Check the options.
Option (1) says all potentials are equal. That is wrong, because \( V \) depends on the wavelength.
Option (2), \( V_2 > V_1 > V_3 \), puts \( V_2 \) above \( V_1 \). That is wrong, since \( \lambda_2 > \lambda_1 \) gives \( V_2 < V_1 \).
Option (4), \( V_3 > V_1 > V_2 \), puts \( V_3 \) above \( V_2 \) and \( V_1 \). That is wrong, since \( \lambda_3 \) is the longest wavelength and so gives the smallest \( V \).

Final Answer:
The shortest wavelength gives the highest stopping potential, so \( V_1 > V_2 > V_3 \), which is option (3). \[ \boxed{V_1 > V_2 > V_3} \]