Question

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and the stopping potential is 2V, what is the wavelength of the electromagnetic wave? Given \( hc = 1242 \, \text{eV} \cdot \text{nm} \) where \( h \) is the Planck constant and \( c \) is the speed of light in vacuum.

• A. 400 nm
• B. 600 nm
• C. 200 nm
• D. 300 nm

Hint:

The energy of the photon in the photoelectric effect can be calculated using the equation \( E = \frac{hc}{\lambda} \), where \( \lambda \) is the wavelength.

Answer 1

The Correct Option is D

The energy of the incident photon is given by: \[ E_{\text{photon}} = \frac{hc}{\lambda}. \] The total energy of the photon is used to overcome the work function \( \phi \) of the metal and to provide the kinetic energy to the ejected electrons. Thus: \[ E_{\text{photon}} = \phi + K.E. \] Substitute the values: \[ \frac{hc}{\lambda} = 2.14 \, \text{eV} + 2 \, \text{eV} = 4.14 \, \text{eV}. \] Using \( hc = 1242 \, \text{eV} \cdot \text{nm} \), we get: \[ \frac{1242}{\lambda} = 4.14 \quad \Rightarrow \quad \lambda = \frac{1242}{4.14} \approx 300 \, \text{nm}. \]