Which figure shows the correct variation of applied potential difference (V) with photoelectric current (I) at two different intensities of light (I₁<I₂) of same wavelengths :

Solution

Physics sir · 3 steps

Same wavelength fixes the stopping potential, and brighter light only lifts the saturation current.

1. Stopping potential. The fastest photoelectron has Kₘₐₓ = hν - φ, and eV₀ = Kₘₐₓ. Both beams have the same wavelength, so V₀ is the same. Both curves must meet the V-axis at the same point -V₀.
2. Saturation current. Intensity sets how many photons arrive each second, so it sets how many electrons are emitted. Since I₁ < I₂, the I₂ curve must stay above the I₁ curve at every positive potential.

   Option	Same cut-off -V₀?	I₂ curve above I₁?
   (a)	No, I₂ starts at V = 0	No, the curves cross
   (b)	No, I₂ cuts off before -V₀	No, the curves cross
   (c)	Yes, both start at -V₀	Yes, everywhere
   (d)	No, I₁ cuts off before -V₀	Yes

3. So. Only (c) has both curves starting from the same -V₀ with the I₂ curve higher throughout.

The idea to keep. Frequency decides the stopping potential, and intensity decides the saturation current. Change only the intensity and the curves share the same cut-off but rise to different heights.

Answer from the NTA final answer key. Solution by PrepWiser.