Question

The following diagram shows the kinetic energy of the ejected photoelectrons against the energy of incident radiation for two metal surfaces M₁ and M₂. If the energy of the incident radiation on M₁ is equal to the work function of M₂, the de Broglie wavelength of the ejected photoelectron is nm

[Given: Mass of electron = 9.11x10⁻³¹ kg; Planck’s constant = 6.62x10⁻³⁴ J s; 1 eV=1.6x10⁻¹⁹ J.]
(round off to two decimal places)

Answer 1

Correct Answer: 1.58 - 1.59

To determine the de Broglie wavelength of the ejected photoelectron, we need to follow these steps:

1. Understanding the Problem:
   The diagram indicates that the threshold energy (work function) for M₁ is 2.3 eV and for M₂ is 2.9 eV. Given: Energy of incident radiation on M₁ equals the work function of M₂, which is 2.9 eV.
2. Calculate Kinetic Energy:
   The kinetic energy (K.E.) of the ejected electron from M₁ is calculated as:
   K.E. = Energy of incident radiation - Work function of M₁
   K.E. = 2.9 eV - 2.3 eV = 0.6 eV
   Convert K.E. to joules: 0.6 eV = 0.6 x 1.6 x 10^-19 J = 9.6 x 10^-20 J
3. Calculate Velocity of Electron:
   Use the kinetic energy formula:
   \( \text{K.E.} = \frac{1}{2}mv^2 \)
   Solving for velocity (v), we have:
   \( v = \sqrt{\frac{2 \times \text{K.E.}}{m}} \)
   \( v = \sqrt{\frac{2 \times 9.6 \times 10^{-20}}{9.11 \times 10^{-31}}} \)
   \( v \approx 4.6 \times 10^5 \, \text{m/s} \)
4. Calculate de Broglie Wavelength:
   The de Broglie wavelength (λ) is given by:
   \( λ = \frac{h}{mv} \)
   where \( h = 6.62 \times 10^{-34} \, \text{J s} \)
   \( λ = \frac{6.62 \times 10^{-34}}{9.11 \times 10^{-31} \times 4.6 \times 10^5} \)
   \( λ \approx 1.58 \times 10^{-9} \, \text{m} \)
5. Verification and Conclusion:
   Convert to nanometers: \( λ \approx 1.58 \, \text{nm} \)

De Broglie Wavelength of the ejected photoelectron: 1.58 nm