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

• A. \( 2K_1 = K_2 \)
• B. \( K_1 = 2K_2 \)
• C. \( K_1<K_2 / 2 \)
• D. \( K_1>2K_2 \)

Hint:

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

Answer 1

The Correct Option is C

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: \[ \text{(C) } K_1<K_2 / 2 \]