Question

The maximum velocity of electrons emitted from a metal surface is \(v\), when frequency of light falling on it is \(f\). The maximum velocity when frequency becomes \(4f\) is

• A. \(2v\)
• B. \(> 2v\)
• C. \(< 2v\)
• D. between \(2v\) and \(4v\)

Hint:

Velocity depends on square root of kinetic energy. Since \(K\) increases more than 4 times, speed increases more than 2 times.

Answer 1

The Correct Option is B

Step 1: Use relation for kinetic energy.
\[ K_{max} = \frac{1}{2}mv^2 = h f - \phi \] Step 2: For frequency \(f\).
\[ \frac{1}{2}mv^2 = hf - \phi \] Step 3: For frequency \(4f\).
\[ \frac{1}{2}mv'^2 = 4hf - \phi \] Step 4: Compare \(v'\) with \(v\).
\[ \frac{v'^2}{v^2} = \frac{4hf - \phi}{hf - \phi} \] Since \(\phi>0\), denominator is smaller than numerator scale, so ratio becomes greater than 4.
Thus,
\[ v'>2v \] Final Answer: \[ \boxed{v'>2v} \]