Given below are two statements :
Statement I : Figure shows the variation of stopping potential with frequency (v) for the two photosensitive materials M₁ and M₂. The slope gives value of h/e, where h is Planck's constant, e is the charge of electron.
Statement II : M₂ will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency.
In the light of the above statements, choose the most appropriate answer from the options given below.

Solution

Physics sir · 4 steps

Every metal gives the same slope h/e; the line further right has the bigger work function.

1. Einstein's equation. The most energetic photoelectron has kinetic energy eV₀ = hν - φ, where φ is the work function. Dividing by e gives a straight line of V₀ against ν.

   V₀ = h/e ν - φ/e

2. Slope check. The slope is h/e, a ratio of universal constants, so it does not depend on the metal. That is why both lines in the figure are parallel. Statement I is correct.
3. Intercept check. Each line meets the ν-axis at the threshold frequency ν₀ = φ/h. M₂ cuts the axis further right, so ν₀ and φ are larger for M₂. At the same ν, hν - φ is smaller for M₂, so its photoelectrons have less kinetic energy. Statement II is incorrect.
4. So. Statement I is correct and Statement II is incorrect, which is option (a).

The idea to keep. On a V₀ against ν graph, every metal has the same slope h/e. The line further right has the larger work function, so at a fixed frequency it gives lower V₀ and lower kinetic energy.

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