Question

Light of wavelength $ 400 \, \text{nm} $ falls on a metal surface with a work function of $ 2.0 \, \text{eV} $. (Planck’s constant $ h = 6.63 \times 10^{-34} \, \text{Js}, \, c = 3 \times 10^8 \, \text{m/s}, \, 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $) Find the maximum kinetic energy of emitted photoelectrons.

• A. 1.1 eV
• B. 2.1 eV
• C. 0.1 eV
• D. 0.8 eV

Hint:

The photoelectric equation is: \[ K.E._{\text{max}} = \frac{hc}{\lambda} - \phi \] Make sure energy is in eV for direct subtraction.

Answer 1

The Correct Option is A

The maximum kinetic energy of emitted photoelectrons is determined by the photoelectric equation: \(K_{\text{max}} = E_{\text{photon}} - \phi\), where \(K_{\text{max}}\) represents the maximum kinetic energy, \(E_{\text{photon}}\) is the energy of the incident photon, and \(\phi\) is the work function of the metal.
Step 1: Calculate \(E_{\text{photon}}\) 
The photon energy is calculated using \(E_{\text{photon}} = \frac{hc}{\lambda}\), with constants \(h = 6.63 \times 10^{-34} \, \text{Js}\), \(c = 3 \times 10^8 \, \text{m/s}\), and wavelength \(\lambda = 400\, \text{nm} = 400 \times 10^{-9} \, \text{m}\).
\(E_{\text{photon}} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \, \text{J}\)
Converting to electron volts (eV) with \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\(E_{\text{photon}} = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.11 \, \text{eV}\)
Step 2: Calculate \(K_{\text{max}}\)
Given a work function \(\phi = 2.0 \, \text{eV}\), the maximum kinetic energy is:
\(K_{\text{max}} = 3.11 \, \text{eV} - 2.0 \, \text{eV} = 1.11 \, \text{eV}\)
The closest option being 1.1 eV, the maximum kinetic energy of the emitted photoelectrons is approximately 1.1 eV.