Question

In a photoelectric effect experiment, light of wavelength \( \lambda \), \( \lambda/2 \), and \( \lambda/6 \) are incident on a metal surface. The stopping potential for these wavelengths are given as \( V_1 \), \( V_2 \), and \( V_3 \), respectively. If the work function of the metal is \( \phi \), calculate the work function using the given wavelengths. The photoelectric equation is given by: \[ E_k = h \nu - \phi \] where:
\( E_k \) is the kinetic energy of the emitted electrons (which is related to the stopping potential),
\( h \) is Planck's constant,
\( \nu \) is the frequency of the incident light,
\( \phi \) is the work function of the metal.

The frequency \( \nu \) is related to the wavelength \( \lambda \) by the equation: \[ \nu = \frac{c}{\lambda} \] where \( c \) is the speed of light.

• A. \( \phi = \frac{h c}{\lambda} \)
• B. \( \phi = \frac{h c}{2 \lambda} \)
• C. \( \phi = \frac{h c}{6 \lambda} \)
• D. \( \phi = \frac{h c}{\lambda} + \frac{h c}{2 \lambda} + \frac{h c}{6 \lambda} \)

Hint:

In photoelectric effect problems, use the relationship between frequency and wavelength and apply the photoelectric equation for each wavelength to find the work function.

Answer 1

The Correct Option is D

The photoelectric equation for each wavelength is: 1. For \( \lambda \): \[ E_k = e V_1 = h \frac{c}{\lambda} - \phi \] 2. For \( \lambda/2 \): \[ E_k = e V_2 = h \frac{c}{\lambda/2} - \phi \] 3. For \( \lambda/6 \): \[ E_k = e V_3 = h \frac{c}{\lambda/6} - \phi \] We can use these equations to find \( \phi \). Adding these gives the work function: \[ \phi = \frac{h c}{\lambda} + \frac{h c}{2 \lambda} + \frac{h c}{6 \lambda} \] Thus, the correct answer is Option (D): \( \phi = \frac{h c}{\lambda} + \frac{h c}{2 \lambda} + \frac{h c}{6 \lambda} \).