Question

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v₁. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v₂. If v₂ = x v₁, the value of x will be ________.

Answer 1

Correct Answer: 2

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,
\(\frac{V_1}{V_2}=\frac{1}{2}\)