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

Step 1: Apply Einstein’s Photoelectric Equation

Einstein’s photoelectric equation states:

\( K_{\text{max}} = hf - hf_0 \)

where \( K_{\text{max}} \) is the maximum kinetic energy of the emitted photoelectrons, \( h \) is Planck’s constant, \( f \) is the frequency of the incident light, and \( f_0 \) is the threshold frequency.

Step 2: Relate Kinetic Energy and Velocity

The maximum kinetic energy is also given by:

\( K_{\text{max}} = \frac{1}{2}mv^2 \)

where \( m \) is the mass of the electron and \( v \) is its velocity.

Step 3: Calculate \( v_1 \)

When \( f = 2f_0 \):

\( \frac{1}{2}mv_1^2 = h(2f_0) - hf_0 = hf_0 \)

Step 4: Calculate \( v_2 \)

When \( f = 5f_0 \):

\( \frac{1}{2}mv_2^2 = h(5f_0) - hf_0 = 4hf_0 \)

Step 5: Find the Ratio \( v_1 / v_2 \)

Dividing the equation for \( v_1^2 \) by the equation for \( v_2^2 \), we get:

\( \frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{hf_0}{4hf_0} \)

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

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

Conclusion: The ratio of \( v_1 \) to \( v_2 \) is \( \frac{1}{2} \) (Option 4).