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 eh, 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.
(A)Statement-I is correct and Statement-II is incorrect.✓ Correct
(B)Statement-I is incorrect but Statement-II is correct.
(C)Both Statement-I and Statement-II are incorrect.
(D)Both Statement-I and Statement-II are correct.
Answer: (A)
Solution
The photoelectric relation is eV0=hν−ϕ, so V0=ehν−eϕ. The slope of V0 vs ν equals eh, common to both materials, so Statement-I is correct. From the graph M2 has the larger threshold frequency, hence the larger work function, so for the same incident frequency M2 emits electrons of lower (not greater) kinetic energy. Statement-II is incorrect.
JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Single correct option
