Question

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v 1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v 2. If v 2 = x v 1, the value of x will be ________.

Answer 1

We know that,
K.E.=hν−ϕ0
where ϕ can be written as hν0
When ν=2ν0
K.E.1=$\frac{mv_1^2}{2}$=2hν0−hν0
V1=$\frac{\sqrt{2h\nu _0}}{m}$
Similarly we can find that
V2=$\frac{\sqrt{4h\nu _0}}{m}$
Hence,
$\frac{V_1}{V_2}=\frac{1}{2}$