Question

Wavelength \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\) are incident on two identical metal plates and photo electrons are emitted. If \(\lambda_{\mathrm{A}}=2 \lambda_{\mathrm{B}}\), the maximum kinetic energy of photo electrons is \(\ldots \ldots \ldots\) (A) \(2 \mathrm{~K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}\) (B) \(\mathrm{K}_{\mathrm{A}}<\left(\mathrm{K}_{\mathrm{B}} / 2\right)\) (C) \(\mathrm{K}_{\mathrm{A}}=2 \mathrm{~K}_{\mathrm{B}}\) (D) \(\mathrm{K}_{\mathrm{A}}>\left(\mathrm{K}_{\mathrm{B}} / 2\right)\)

Short answer

(B) \(K_{A}< \left(K_{B} / 2\right)\)

Step-by-step solution

Recall the photoelectric effect equation

The photoelectric effect equation is given by: \[E_g = K_{max} + W\] where \(E_g\) is the energy of the incident photon, \(K_{max}\) is the maximum kinetic energy of the emitted photoelectron, and \(W\) is the work function of the metal plate (minimum energy required to release a photoelectron).

Recall the relationship between wavelength and energy

The energy of a photon is related to its wavelength (\(\lambda\)) by the Planck's equation: \[E_g = \frac{hc}{\lambda}\] where \(h\) is the Planck's constant and \(c\) is the speed of light.

Write the equations for the two photons

Using the above information, we can write the energy equations for the two incident lights with wavelengths \(\lambda_A\) and \(\lambda_B\): \[E_{gA} = \frac{hc}{\lambda_A} = K_{maxA} + W\] \[E_{gB} = \frac{hc}{\lambda_B} = K_{maxB} + W\]

Use the given relationship between the wavelengths

We are given that \(\lambda_A = 2\lambda_B\). Substitute this relationship in the equation for \(E_{gA}\): \[E_{gA} = \frac{hc}{2\lambda_B} = K_{maxA} + W\] Now solve for \(K_{maxA}\): \[K_{maxA} = \frac{hc}{2\lambda_B} - W\]

Compare the maximum kinetic energies

Let's compare \(K_{maxA}\) and \(K_{maxB}\): \[\frac{K_{maxA}}{K_{maxB}} = \frac{\frac{hc}{2\lambda_B} - W}{\frac{hc}{\lambda_B} - W}\] We want to find the condition that compares \(K_{maxA}\) and \(K_{maxB}\) to match one of the given options. Simplify the above expression: \[\frac{K_{maxA}}{K_{maxB}} = \frac{\frac{1}{2} - \frac{W\lambda_B}{hc}}{1 - \frac{W\lambda_B}{hc}}\]

Analyze the obtained expression

The fraction on the right side of the expression cannot exceed 1, as the work function (\(W\)) and the wavelength (\(\lambda_B\)) are both positive: \[\frac{W\lambda_B}{hc} \le 1\] Since \(K_{maxA}\) is multiplied by \(\frac{1}{2}\) on the numerator, the fraction on the left-hand side will always be less than 1: \[\frac{K_{maxA}}{K_{maxB}} < 1\] This means that the maximum kinetic energy of photoelectrons due to the incident light with wavelength \(\lambda_A\) is less than that due to the incident light with wavelength \(\lambda_B\): \[K_{maxA} < K_{maxB}\] Comparing our result with the given options, we can conclude that the correct answer is: (B) \(K_{A}< \left(K_{B} / 2\right)\)