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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

If de-Broglie wavelength of electron is increased by \(1 \%\) its momentum \(\ldots \ldots\) (A) increases by \(1 \%\) (B) decreases by \(1 \%\) (C) increased by \(2 \%\) (D) decreases by \(2 \%\)

Work function of \(\mathrm{Zn}\) is \(3.74 \mathrm{eV}\). If the sphere of \(\mathrm{Zn}\) is illuminated by the X-ray of wavelength \(12 \AA\) the maximum potential produced on the sphere is \(\ldots \ldots \ldots\) \(\left(\mathrm{h}-6.625 \times 10^{-34 \mathrm{~J}-\mathrm{s}}\right)\) (A) \(10.314 \mathrm{~V}\) (B) \(103.14 \mathrm{~V}\) (C) \(1031.4 \mathrm{~V}\) (D) \(10314 \mathrm{~V}\)

Suppose \(\Psi(\mathrm{x}, \mathrm{y}, \mathrm{z})\) represents a particle in three dimensional space, then probability of finding the particle in the unit volume at a given point \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) is $\ldots \ldots$ (A) inversely proportional to $\Psi^{\prime}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ (B) directly proportional \(\Psi^{*}\) (C) directly proportional to \(\mid \Psi \Psi^{*}\) (D) inversely proportional to \(\left|\Psi \Psi^{*}\right|\)

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The cathode of a photoelectric cell is changed such that the work function changes from \(\mathrm{W}_{1}\) to \(\mathrm{W}_{2}\left(\mathrm{~W}_{2}>\mathrm{W}_{1}\right)\). If the currents before and after change are \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\), all other conditions remaining unchanged, then assuming $\mathrm{hf}>\mathrm{W}_{2} \ldots \ldots$ (A) \(\mathrm{I}_{1}=\mathrm{I}_{2}\) (B) \(I_{1}\mathrm{I}_{2}\) (D) \(\mathrm{I}_{1}<\mathrm{I}_{2}<2 \mathrm{I}_{1}\)
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