Question

When a metal surface is illuminated by a light of wavelength \( \lambda \), the stopping potential is \( V \). If the same surface is illuminated by light of wavelength \( 2\lambda \), the stopping potential is \( \frac{V}{4} \), the threshold wavelength is:

• A. \( \lambda \)
• B. \( 2\lambda \)
• C. \( 3\lambda \)
• D. \( \frac{\lambda}{2} \)

Hint:

Remember that doubling the wavelength in the context of the photoelectric effect will change the energy of the photons and thus the stopping potential, but the threshold wavelength is a material constant.

Answer 1

The Correct Option is C

The photoelectric effect is described by Einstein's photoelectric equation:

\[ K_{\text{max}} = h \nu - \phi \]

where:

• \( K_{\text{max}} \) is the maximum kinetic energy of the emitted electrons,
• \( h \) is Planck's constant,
• \( \nu \) is the frequency of the incident light,
• \( \phi \) is the work function of the metal.

The stopping potential \( V \) is related to the maximum kinetic energy by:

\[ K_{\text{max}} = eV \]

Thus, we can write the equation for stopping potential as:

\[ eV = h \nu - \phi \]

We are given that the stopping potential is \( V \) when the wavelength of the incident light is \( \lambda \). Therefore, the energy equation for this situation is:

\[ eV = h \frac{c}{\lambda} - \phi \]

where \( c \) is the speed of light.

Now, when the wavelength of the light is \( 2\lambda \), the stopping potential is \( \frac{V}{4} \). The energy equation for this situation becomes:

\[ e \frac{V}{4} = h \frac{c}{2\lambda} - \phi \] Step 1: Subtract the two equations

Now, we subtract the second equation from the first to eliminate the work function \( \phi \):

\[ eV - e \frac{V}{4} = h \left( \frac{c}{\lambda} - \frac{c}{2\lambda} \right) \]

Simplifying:

\[ \frac{3eV}{4} = h \left( \frac{c}{2\lambda} \right) \] \[ \frac{3eV}{4} = \frac{hc}{2\lambda} \] Step 2: Solve for \( V \)

Now, we can solve for the wavelength \( \lambda \):

\[ V = \frac{2\lambda e}{hc} \times 3 \]

Thus, the threshold wavelength is \( 3\lambda \).

Thus, the correct answer is Option (3), \( 3\lambda \).