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Chapter 17: Problem 2378

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

Expert verified
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

01

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
02

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
03

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
04

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: \[\lambda = \frac{h}{\sqrt{2mq_1V}}\]
05

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: \(h = 6.63 × 10^{-34} \mathrm{~J s}\) \(m = 9.11 \times 10^{-31} \mathrm{~kg}\) \(\lambda = 0.4 \AA = 0.4 \times 10^{-10}\mathrm{~m}\) \(q_1 = 1.6 \times 10^{-19}\mathrm{~C}\) \[V = \frac{(6.63 \times 10^{-34})^2}{2(9.11 \times 10^{-31})(0.4 \times 10^{-10})^2(1.6 \times 10^{-19})}\]
06

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) \( 94.10 \mathrm{~V} \)

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Most popular questions from this chapter

Ration of momentum of photons having wavelength \(4000 \AA \& 8000 \AA\) is ........... (A) \(2: 1\) (B) \(1: 2\) (C) \(20: 1\) (D) \(1: 20\)

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A proton, a deuteron and an \(\propto\) -particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is (A) \(1: 2: 8\) (B) \(1: 2: 4\) (C) \(1: 1: 2\) (D) None of these

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Frequency of incident light on body is \(\mathrm{f}\). Threshold frequency of body is \(f_{0}\). Maximum velocity of electron \(=\ldots \ldots \ldots\).. where \(m\) is mass of electron. (A) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]^{(1 / 2)}$ (B) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]$ (C) \([2 \mathrm{hf} / \mathrm{m}]^{(1 / 2)}\) (D) \(\mathrm{h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\)
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