Question:

Let $ K_1 $ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $ \lambda_1 $ and $ K_2 $ corresponding to wavelength $ \lambda_2 $. If $ \lambda_1 = 2\lambda_2 $, then

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The kinetic energy of the photoelectrons is inversely proportional to the wavelength of the light. As the wavelength increases, the kinetic energy decreases.

Updated On: May 3, 2025

$ 2K_1 = K_2 $
$ K_1 = 2K_2 $
$ K_1<K_2 / 2 $
$ K_1>2K_2 $

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The Correct Option is C

Approach Solution - 1

The maximum kinetic energy of the photoelectrons emitted is related to the wavelength of light by Einstein's photoelectric equation: \[ K = h\nu - \phi \] where:
- $ K $ is the maximum kinetic energy of the photoelectrons,
- $ h $ is Planck's constant,
- $ \nu $ is the frequency of the incident light,
- $ \phi $ is the work function of the material. Since the frequency $ \nu $ is related to the wavelength $ \lambda $ by the equation $ \nu = \frac{c}{\lambda} $, where $ c $ is the speed of light, we can write the kinetic energy as: \[ K = h \left( \frac{c}{\lambda} \right) - \phi \] Now, let’s compare the kinetic energies for the two wavelengths: - For $ \lambda_1 $, the maximum kinetic energy is $ K_1 = h \left( \frac{c}{\lambda_1} \right) - \phi $, - For $ \lambda_2 $, the maximum kinetic energy is $ K_2 = h \left( \frac{c}{\lambda_2} \right) - \phi $. Given that $ \lambda_1 = 2\lambda_2 $, we have: \[ K_1 = h \left( \frac{c}{2\lambda_2} \right) - \phi \] \[ K_2 = h \left( \frac{c}{\lambda_2} \right) - \phi \] From the equations above, we can see that $ K_1 $ is less than $ K_2 / 2 $. This is because the frequency of the light corresponding to $ \lambda_1 $ is half of the frequency corresponding to $ \lambda_2 $, leading to a smaller kinetic energy for $ \lambda_1 $.
Thus, the correct answer is: \[ \text{(C) } K_1<K_2 / 2 \]

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Approach Solution -2

The photoelectric effect describes the phenomenon where light incident on a material causes it to emit electrons. The maximum kinetic energy ($ K $) of these photoelectrons is determined by the equation:

$ K = h\nu - \phi $

where:

$ h $ is Planck's constant.
$ \nu $ is the frequency of the incident light.
$ \phi $ is the work function of the material.

The frequency ($ \nu $) is related to the wavelength ($ \lambda $) of light by:

$ \nu = \frac{c}{\lambda} $

where $ c $ is the speed of light. Therefore, the kinetic energy can be expressed as:

$ K = \frac{hc}{\lambda} - \phi $

Given two wavelengths, $ \lambda_1 $ and $ \lambda_2 $, the kinetic energies $ K_1 $ and $ K_2 $ for the corresponding wavelengths are:

$ K_1 = \frac{hc}{\lambda_1} - \phi $

$ K_2 = \frac{hc}{\lambda_2} - \phi $

According to the problem, $ \lambda_1 = 2\lambda_2 $. Substitute this relation into the kinetic energy formula:

$ K_1 = \frac{hc}{2\lambda_2} - \phi $

Comparing $ K_1 $ and $ K_2 $:

$ K_2 = \frac{hc}{\lambda_2} - \phi $

Now, substitute $ K_1 $ in terms of $ K_2 $:

$ K_1 = \frac{1}{2} \left(\frac{hc}{\lambda_2} - \phi\right) = \frac{1}{2} (K_2 + \phi) - \phi = \frac{1}{2} K_2 - \frac{1}{2} \phi $

Since $\phi$ is positive, it follows that:

$ K_1 < \frac{1}{2} K_2 $

Thus, the correct answer is: $ K_1 < \frac{K_2}{2} $.