Khimraj
Last Activity: 6 Years ago
The photoelectric eqn is:
hf - φ = qVs,
where, the stopping potential can be solved;
Vs = (hf - φ)/q
let f' = 2f
and work function: φ of the metal is constant.
∆Vs = Vs' - Vs = [(h*2f - φ) - (hf - φ)]/q = hf/q
while,
2Vs = 2(hf - φ))/q
so that,
∆Vs > 2Vs
meaning that if the frequency is doubled the
stopping potential, Vs increases to more than doubled.
