The threshold frequency of metal is \(f_0\). When the light of frequency \(2 f_0\) is incident on the metal plate, the maximum velocity of photoelectron is \(v_1\). When the frequency of incident radiation is increased to \(5 f _0\). the maximum velocity of photoelectrons emitted is \(v_2\). The ratio of \(v_1\) to \(v_2\) is:

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Answer & Solution

Correct Answer

(A) \(\frac{v_1}{v_2}=\frac{1}{2}\)

Detailed explanation

\(K _{\max }= hf - hf _0\)

For \(f =2 f _0\)

\(\frac{1}{2} m V _1^2=2 h f _0- hf _0= hf f _0\)

For \(f =5 f _0\)

\(\frac{1}{2} m V _2^2=5 h \textrm {f } _ { 0 } - hf _0=4 h _0\)

\(\frac{ V _1}{ V _2}=\frac{1}{2}\)

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