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

To find the maximum kinetic energy of the emitted photoelectrons, we use the photoelectric equation: \(K_{\text{max}} = E_{\text{photon}} - \phi\), where \(K_{\text{max}}\) is the maximum kinetic energy, \(E_{\text{photon}}\) is the energy of the incoming photon, and \(\phi\) is the work function of the metal. 
Step 1: Calculate \(E_{\text{photon}}\) 
The energy of a photon can be calculated using the equation \(E_{\text{photon}} = \frac{hc}{\lambda}\), where \(h = 6.63 \times 10^{-34} \, \text{Js}\), \(c = 3 \times 10^8 \, \text{m/s}\), and \(\lambda = 400\, \text{nm} = 400 \times 10^{-9} \, \text{m}\).
\(E_{\text{photon}} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \, \text{J}\)
Convert this energy to electron volts (eV) using the conversion \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\(E_{\text{photon}} = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.11 \, \text{eV}\)
Step 2: Calculate \(K_{\text{max}}\)
Given \(\phi = 2.0 \, \text{eV}\), the maximum kinetic energy is:
\(K_{\text{max}} = 3.11 \, \text{eV} - 2.0 \, \text{eV} = 1.11 \, \text{eV}\)
Since the closest option is 1.1 eV, the maximum kinetic energy of the emitted photoelectrons is approximately 1.1 eV.

Answer 2

Energy of incident photon: \[ E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \, \text{J} = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \, \text{J} \] Convert to eV: \[ E = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.1 \, \text{eV} \] Work function \( \phi = 2.0 \, \text{eV} \) Maximum K.E. of emitted photoelectron: \[ K.E. = E - \phi = 3.1 - 2.0 = \boxed{1.1 \, \text{eV}} \]