Question

The threshold frequency of a metal corresponds to the wavelength of \(x\) nm. In two separate experiments ‘A’ and ‘B’, incident radiations of wavelengths \(\frac{x}{2}\) nm and \(\frac{x}{4}\) nm respectively are used. The ratio of kinetic energy of the released electrons in experiment ‘B’ to that in experiment ‘A’ is

• A. \(\frac{1}{3}\)
• B. \(2\)
• C. \(4\)
• D. \(3\)
• E. \(\frac{1}{2}\)

Hint:

Always convert wavelength into reciprocal form for photoelectric problems.

Answer 1

The Correct Option is D

Concept: Photoelectric equation: \[ K_{\max} = h\nu - h\nu_0 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) \]

Step 1: Threshold wavelength. \[ \lambda_0 = x \]

Step 2: Kinetic energy in experiment A. \[ \lambda_A = \frac{x}{2} \] \[ K_A = hc\left(\frac{1}{x/2} - \frac{1}{x}\right) \] \[ = hc\left(\frac{2}{x} - \frac{1}{x}\right) = \frac{hc}{x} \]

Step 3: Kinetic energy in experiment B. \[ \lambda_B = \frac{x}{4} \] \[ K_B = hc\left(\frac{1}{x/4} - \frac{1}{x}\right) \] \[ = hc\left(\frac{4}{x} - \frac{1}{x}\right) = \frac{3hc}{x} \]

Step 4: Ratio. \[ \frac{K_B}{K_A} = \frac{3hc/x}{hc/x} = 3 \]

Step 5: Final answer. \[ \boxed{3} \]