If the frequency of light in a photoelectric experiment is doubled the stopping potential will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → becomes more than the double
Einstein's photoelectric equation relates the stopping potential \( V_0 \) to the frequency \( \nu \) and the work function \( \phi_0 \):
\( \displaystyle eV_0 = h\nu - \phi_0 \)
With the frequency doubled:
\( \displaystyle eV_0' = 2h\nu - \phi_0 = 2(h\nu - \phi_0) + \phi_0 = 2eV_0 + \phi_0 \)
\( \displaystyle V_0' = 2V_0 + \frac{\phi_0}{e} \gt 2V_0 \)
So the stopping potential becomes more than double.
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Using Eq. (11.1), the photoelectric equation, Eq. (11.2), can be written as
\( \displaystyle eV_0 = h\nu - \phi_0; \quad \text{for} \quad \nu \geq \nu_0 \)
\( \displaystyle \text{or} \quad V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \qquad (11.4) \)
This is an important result. It predicts that the V0 versus ν curve is a straight line with slope = (h/e), independent of the nature of the material.”