When two monochromatic lights of frequency u and u/2 are incident on a photoelectric metal, their stopping potentials become (Vₛ)/2 and Vₛ respectively. The threshold frequency for this metal is:

Question

When two monochromatic lights of frequency 
u and u/2 are incident on a photoelectric metal, their stopping potentials become (Vₛ)/2 and Vₛ respectively. The threshold frequency for this metal is:

Accepted Answer

(4) 3/2
u. Write Einstein's equation once for each light, in the form eV₀=h
u-h
u₀. At frequency 
u the stopping potential is (Vₛ)/2, so (eVₛ)/2=h
u-h
u₀. At frequency u/2 the stopping potential is Vₛ, so eVₛ=(h
u)/2-h
u₀. Doubling the first equation gives eVₛ=2h
u-2h
u₀, and this can be put into the second. 2h
u-2h
u₀=(h
u)/2-h
u₀, so 2h
u-(h
u)/2=2h
u₀-h
u₀. (3h
u)/2=h
u₀, giving 
u₀=(3
u)/2. One caution is worth carrying away: this threshold is higher than either incident frequency, so on these numbers neither light would eject an electron at all. The data as printed are not self-consistent, and the value above is what the paper's own algebra gives and what its key marks.

When two monochromatic lights of frequency u and u/2 are incident on a photoelectric metal, their stopping potentials become (Vₛ)/2 and Vₛ respectively. The threshold frequency for this metal is:

Step-by-step solution

Why the other options are wrong