Question

The threshold frequency of a metal is \(f _0\) When the light of frequency \(2 f _0\) is incident on the metal plate, the maximum velocity of photoelectrons 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 :

• A. \(\frac{v_1}{v_2}=\frac{1}{8}\)
• B. \(\frac{v_1}{v_2}=\frac{1}{4}\)
• C. \(\frac{v_1}{v_2}=\frac{1}{16}\)
• D. \(\frac{v_1}{v_2}=\frac{1}{2}\)

Hint:

Remember Einstein’s photoelectric equation and how to relate the maximum kinetic energy of photoelectrons to their velocity.

Answer 1

The Correct Option is D

To solve this problem, let's use the photoelectric effect equation. According to Einstein's photoelectric equation:

\(E_k = hf - \phi\)

where:

• \(E_k\) is the kinetic energy of the photoelectrons.
• \(h\) is Planck's constant.
• \(f\) is the frequency of incident light.
• \(\phi = hf_0\) is the work function of the metal, with \(f_0\) being the threshold frequency.

From the kinetic energy, the maximum velocity \(v\) of the emitted photoelectrons is given by:

\(E_k = \frac{1}{2} mv^2\)

For frequency \(2f_0\):

\(E_{k1} = h(2f_0) - hf_0 = hf_0\)

This gives:

\(\frac{1}{2}mv_1^2 = hf_0\)

For frequency \(5f_0\):

\(E_{k2} = h(5f_0) - hf_0 = 4hf_0\)

This gives:

\(\frac{1}{2}mv_2^2 = 4hf_0\)

Let us find the ratio \(\frac{v_1}{v_2}\):

\(\frac{v_1^2}{v_2^2} = \frac{hf_0 / \frac{1}{2}m}{4hf_0 / \frac{1}{2}m} = \frac{1}{4}\)

Taking the square root on both sides:

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

Hence, the correct option is \(\frac{v_1}{v_2}=\frac{1}{2}\).