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Chapter 17: Problem 2421

Through what potential difference should an electron be accelerated so it's de-Broglie wavelength is \(0.3 \AA\). (A) \(1812 \mathrm{~V}\) (B) \(167.2 \mathrm{~V}\) (C) \(1516 \mathrm{~V}\) (D) \(1672.8 \mathrm{~V}\)

Short Answer

Expert verified
To find the potential difference through which an electron must be accelerated for its de Broglie wavelength to be 0.3 Å, we can use the combined formula \(V = \frac{h^2}{2m_eqλ^2}\). Plugging in the values for Planck's constant (h), electron's mass (\(m_e\)), charge (q), and the given wavelength (λ), we get \(V \approx 167.2 \, \text{V}\). Therefore, the correct answer is (B) \(167.2 \mathrm{~V}\).

Step by step solution

01

Understand the de Broglie wavelength formula

For any particle, the de Broglie wavelength (λ) is given by the formula: \(λ = \frac{h}{p}\) Where h is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)), and p is the particle's momentum. For an electron, its momentum can be calculated using the formula: \(p = \sqrt{2m_eE}\) Where \(m_e\) is the electron's mass (\(9.11 \times 10^{-31} \, \text{kg}\)), and E is its energy.
02

Relating energy to potential difference

To find the energy of the electron when it is accelerated through a potential difference V, we can use the formula: \(E = qV\) Where E is the energy, q is the charge of the electron (which is \(1.6 \times 10^{-19} \, \text{C}\)), and V is the potential difference.
03

Combine formulas to calculate potential difference

We want to find the potential difference V, so we will combine the formulas for the de Broglie wavelength, momentum, and energy as follows: \(λ = \frac{h}{\sqrt{2m_eE}} = \frac{h}{\sqrt{2m_eqV}}\) Now, we can solve for V: \(V = \frac{h^2}{2m_eqλ^2}\)
04

Calculate the potential difference V

Now that we have the formula for the potential difference V, we can plug in the given values and solve for V: \(V = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})^2}{2(9.11 \times 10^{-31} \, \text{kg})(1.6 \times 10^{-19} \, \text{C})(0.3 \times 10^{-10} \, \text{m})^2}\) After calculating, we get: \(V \approx 167.2 \, \text{V}\) So, the correct answer is: (B) \(167.2 \mathrm{~V}\)

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Most popular questions from this chapter

An electron is accelerated under a potential difference of \(64 \mathrm{~V}\), the de Broglie wave length associated with electron is $=\ldots \ldots \ldots \ldots . \AA$ (Use charge of election $=1.6 \times 10^{-19} \mathrm{C},=9.1 \times 10^{-31} \mathrm{Kg}$, mass of electrum \(\mathrm{h}=6.6623 \times 10^{-43} \mathrm{~J}\).sec \()\) (A) \(4.54\) (B) \(3.53\) (C) \(2.53\) (D) \(1.534\)

Select the correct statement from the following (A) Radiation and matter (particles) may not exhibit both the wave nature and particle nature simultaneously at the some moment. (B) At some moment electromagnetic waves get divided in to small pieces named particles (C) In a given circumstance a particle at one moment behaves like particle and the at the next moment as wave and so on (D) Each microscopic particle is enveloped by a wave

A proton and an \(\propto\) -particle are passed through same potential difference. If their initial velocity is zero, the ratio of their de Broglie's wavelength after getting accelerated is \(\ldots \ldots\) (A) \(1: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(2 \sqrt{2}: 1\)

A \(100 \mathrm{~W}\) bulb radiates energy at rate of $100 \mathrm{~J} / \mathrm{s}\(. If the light emitted has wavelength of \)525 \mathrm{~nm}$. How many photons are emitted per second? (A) \(1.51 \times 10^{20}\) (B) \(1.51 \times 10^{22}\) (C) \(2.64 \times 10^{20}\) (D) \(4.5 \times 10^{19}\)

Wavelength \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\) are incident on two identical metal plates and photo electrons are emitted. If \(\lambda_{\mathrm{A}}=2 \lambda_{\mathrm{B}}\), the maximum kinetic energy of photo electrons is \(\ldots \ldots \ldots\) (A) \(2 \mathrm{~K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}\) (B) \(\mathrm{K}_{\mathrm{A}}<\left(\mathrm{K}_{\mathrm{B}} / 2\right)\) (C) \(\mathrm{K}_{\mathrm{A}}=2 \mathrm{~K}_{\mathrm{B}}\) (D) \(\mathrm{K}_{\mathrm{A}}>\left(\mathrm{K}_{\mathrm{B}} / 2\right)\)
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