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Chapter 17: Problem 2360

If electron is accelerated under \(50 \mathrm{KV}\) in microscope, find its de- Broglie wavelength. (A) \(5.485 \times 10^{-12} \mathrm{~m}\) (B) \(8.545 \times 10^{-12} \mathrm{~m}\) (C) \(4.585 \times 10^{-12} \mathrm{~m}\) (D) \(5.845 \times 10^{-12} \mathrm{~m}\)

Short Answer

Expert verified
The de Broglie wavelength of an electron accelerated under a \(50 \mathrm{kV}\) potential difference in a microscope is approximately \(5.485 \times 10^{-12} \mathrm{m}\), which corresponds to answer choice (A).

Step by step solution

01

Recall the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \[\lambda = \frac{h}{p}\] Where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle. For an electron accelerated under a potential difference, we can use the following formula: \[E = \frac{1}{2}mv^2 = eV\] Where: - \(E\) is the energy of the electron - \(m\) is the mass of the electron - \(v\) is the velocity of the electron - \(e\) is the elementary charge - \(V\) is the potential difference
02

Find the momentum

We know that the momentum is given by: \[p = mv\] Now from the energy formula, we have: \[v = \sqrt{\frac{2eV}{m}}\] Substitute the values of the energy and mass into the momentum formula: \[p = m \cdot \sqrt{\frac{2eV}{m}}\] \[p = \sqrt{2meV}\]
03

Calculate the de Broglie wavelength

Now, we can substitute the momentum formula into the de Broglie wavelength formula: \[\lambda = \frac{h}{\sqrt{2meV}}\] Given values: - Planck's constant (h) = \(6.626 \times 10^{-34} \mathrm{Js}\) - Elementary charge (e) = \(1.6 \times 10^{-19} \mathrm{C}\) - Mass of electron (m) = \(9.11 \times 10^{-31} \mathrm{kg}\) - Potential difference (V) = \(50 \times 10^3 \mathrm{V}\) (converted from kV to V) Substitute the given values into the wavelength formula: \[\lambda = \frac{6.626 \times 10^{-34} \mathrm{Js}}{\sqrt{2 \cdot (9.11 \times 10^{-31} \mathrm{kg}) \cdot (1.6 \times 10^{-19} \mathrm{C}) \cdot (50 \times 10^3 \mathrm{V})}}\]
04

Calculate the result and compare with the answer choices

Perform the calculation: \[\lambda \approx 5.485 \times 10^{-12} \mathrm{m}\] Comparing this result with the given answer choices, we find that the correct answer is (A): \[(A) \; 5.485 \times 10^{-12} \mathrm{m}\]

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Most popular questions from this chapter

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 \(\left|\Psi \Psi^{*}\right|\)

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) \(1.534\)

Work function of metal is \(2.5 \mathrm{eV}\) If wave length of light incident on metal plate is \(3000 \AA\), stopping potential of emitted electron will be....... $\left\\{\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(0.82 \mathrm{~V}\) (B) \(0.41 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(3.28 \mathrm{~V}\)

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Calculate the energy of a photon of radian wavelength \(6000 \AA\) in \(\mathrm{eV}\) (A) \(20.6 \mathrm{eV}\) (B) \(2.06 \mathrm{eV}\) (C) \(1.03 \mathrm{eV}\) (D) \(4.12 \mathrm{eV}\)
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