Question

In a photoelectric experiment, when the wavelength of the incident light on a metal is λ, the maximum kinetic energy of the emitted photoelectron is E. When the wavelength of incident light is \(\frac{\lambda}{3}\), the maximum kinetic energy of the emitted photoelectron becomes 4E. The work function of the metal is:

• A. $\frac{h c}{\lambda}$
• B. $\frac{h c}{3 \lambda}$
• C. $\frac{4 h c}{\lambda}$
• D. $4 E$

Hint:

Use the photoelectric equation $E_{{max}} = \frac{h c}{\lambda} - \phi$. Set up equations for different wavelengths and solve for the work function $\phi$.

Answer 1

The Correct Option is B

Einstein’s photoelectric equation: $E_{{max}} = \frac{h c}{\lambda} - \phi$, where $\phi$ is the work function.
Case 1: Wavelength $\lambda$, $E_{{max}} = E \implies E = \frac{h c}{\lambda} - \phi$ (Equation 1).
Case 2: Wavelength $\frac{\lambda}{3}$, $E_{{max}} = 4E \implies 4E = \frac{h c}{\frac{\lambda}{3}} - \phi = \frac{3 h c}{\lambda} - \phi$ (Equation 2).
Subtract Equation 1 from Equation 2: $4E - E = \left(\frac{3 h c}{\lambda} - \phi\right) - \left(\frac{h c}{\lambda} - \phi\right)$.
Simplify: $3E = \frac{3 h c}{\lambda} - \frac{h c}{\lambda} = \frac{2 h c}{\lambda}$.
So, $E = \frac{2 h c}{3 \lambda}$.
Substitute $E$ into Equation 1: $\frac{2 h c}{3 \lambda} = \frac{h c}{\lambda} - \phi$.
Solve for $\phi$: $\phi = \frac{h c}{\lambda} - \frac{2 h c}{3 \lambda} = \frac{3 h c - 2 h c}{3 \lambda} = \frac{h c}{3 \lambda}$.