Question

With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is \(0.4 \AA\) (A) \(9410 \mathrm{~V}\) (B) \(94.10 \mathrm{~V}\) (C) \(9.140 \mathrm{~V}\) (D) \(941.0 \mathrm{~V}\)

Short answer

The potential difference required to accelerate an electron so that its de-Broglie wavelength is \(0.4 \AA\) is approximately \(94.1 \mathrm{~V}\). Thus, the correct option is (B) \( 94.10 \mathrm{~V} \).

Step-by-step solution

1. Write down the de-Broglie wavelength formula

The de-Broglie wavelength formula is given by: \[\lambda = \frac{h}{mv}\] Where, \( \lambda \) = de-Broglie wavelength \( h \) = Planck’s constant = \(6.63 × 10^{-34} \mathrm{~J s}\) \( m \) = Mass of the electron \( v \) = Velocity of the electron

2. Write down the formula to find the kinetic energy of the electron

The kinetic energy of the electron is given by: \[KE = \frac{1}{2}mv^2\] Where, \( KE \) = Kinetic energy \( m \) = Mass of the electron \( v \) = Velocity of the electron

3. Find the relationship between potential difference, charge, and kinetic energy

The potential difference through which the electron should be accelerated can be determined by the relation: \[KE = qV\] Where, \( KE \) = Kinetic energy \( q \) = Charge of the electron \( V \) = Potential difference

4. Combine the de-Broglie wavelength formula, the kinetic energy formula, and the relation between potential difference, charge, and kinetic energy

The de-Broglie wavelength formula can be combined with the kinetic energy formula, and the potential difference relation as: \[\lambda = \frac{h}{\sqrt{2mq_1V}}\]

5. Solve for the potential difference, V

Rearrange and solve for the potential difference, V: \[V = \frac{h^2}{2m \lambda^2q_1}\] Now, plug in the values for h, m, \(\lambda\), and q_1: \(h = 6.63 × 10^{-34} \mathrm{~J s}\) \(m = 9.11 \times 10^{-31} \mathrm{~kg}\) \(\lambda = 0.4 \AA = 0.4 \times 10^{-10}\mathrm{~m}\) \(q_1 = 1.6 \times 10^{-19}\mathrm{~C}\) \[V = \frac{(6.63 \times 10^{-34})^2}{2(9.11 \times 10^{-31})(0.4 \times 10^{-10})^2(1.6 \times 10^{-19})}\]

6. Calculate the potential difference, V

After plugging in the values, the potential difference V can be calculated: \[V \approx 94.1 \mathrm{~V}\] Hence, the correct option is: (B) \( 94.10 \mathrm{~V} \)