An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is. (nearly): (mt. = $9 \times 10^{-31}$ kg)

Question

The variation of the stopping potential $ V_0 $ with the frequency $ \nu $ of the incident radiation for four metals A, B, C, and D is shown in the figure. For the same frequency of incident radiation producing photo-electrons in all metals, the kinetic energy of photo-electrons will be maximum for which metal?
variation of the stopping potential

Answer 1

To solve the problem of determining the de Broglie wavelength of an electron accelerated through a potential difference of 10,000 V, we can use the de Broglie wavelength formula:

$\lambda = \frac{h}{p}$

where $ \lambda $ is the de Broglie wavelength, $ h $ is Planck's constant $ (6.626 \times 10^{-34} \, \text{Js}) $, and $ p $ is the momentum of the electron.

The momentum $ p $ can also be expressed in terms of the kinetic energy acquired by the electron when accelerated through a potential difference $ V $ (in volts):

$p = \sqrt{2m \cdot e \cdot V}$

where $ m $ is the mass of the electron $ (9 \times 10^{-31} \, \text{kg}) $, and $ e $ is the elementary charge $ (1.6 \times 10^{-19} \, \text{C}) $.

By substituting for $ p $ in the de Broglie wavelength formula, we get:

$\lambda = \frac{h}{\sqrt{2m \cdot e \cdot V}}$

Substituting the given values:

h = 6.626 \times 10^{-34} \, \text{Js}
m = 9 \times 10^{-31} \, \text{kg}
e = 1.6 \times 10^{-19} \, \text{C}
V = 10,000 \, \text{V}

We calculate:

\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10,000}} \] \[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2.88 \times 10^{-15}}} \] \] \[ \lambda \approx \frac{6.626 \times 10^{-34}}{5.366 \times 10^{-8}} \] \] \[ \lambda \approx 12.36 \times 10^{-12} \, \text{m} \]

On rounding, the de Broglie wavelength of the electron is approximately $\textbf{12.2} \times 10^{-12} \, \text{m}$.

Thus, the correct answer is $12.2\times 10^{-12}$m.

Answer 2