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Chapter 17: Problem 2333

If we take accelerating voltage \(\mathrm{V}=50 \mathrm{~V}\), electric charge of electron \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\) and mass of electron \(\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}\) find the wavelength of concerned election. (A) \(0.1735 \mathrm{~A}\) (B) \(1.735 \AA\) (C) \(17.35 \AA\) (D) \(1735 \AA\)

Short Answer

Expert verified
The short answer is (B) \(1.735 \AA\).

Step by step solution

01

Write down 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 electron.
02

Write down the energy relation for an electron in an accelerating voltage

The energy relation for an electron in an accelerating voltage is given by: \[ E = eV \] where \(E\) is the energy of the electron, \(e\) is the electric charge of the electron, and \(V\) is the accelerating voltage.
03

Relate the energy and the momentum of the electron

The energy and the momentum of the electron are related by the following formula: \[ E = \frac{p^2}{2m} \] where \(E\) is the energy, \(p\) is the momentum, and \(m\) is the mass of the electron.
04

Express the momentum in terms of energy, electric charge, and accelerating voltage

Using the energy relation and the momentum-energy relation, we can express the momentum of the electron as follows: \[ p = \sqrt{2m eV} \]
05

Calculate the de Broglie wavelength using the given values

Now, we can substitute the given values and the derived momentum expression into the de Broglie wavelength formula to calculate the wavelength: \[ \lambda = \frac{h}{\sqrt{2m eV}} \] Plugging in the values, \(h = 6.626 \times 10^{-34} \mathrm{Js}\), \(V = 50 \mathrm{~V}\), \(e = 1.6 \times 10^{-19} \mathrm{C}\), and \(m = 9.1 \times 10^{-31} \mathrm{~kg}\): \[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2(9.1 \times 10^{-31})(1.6 \times 10^{-19})(50)}} \]
06

Solve for the wavelength and convert it to Angstroms

Now, solving the above expression, we get the following value for the wavelength in meters: \[ \lambda = 1.735 \times 10^{-10} \mathrm{m} \] To convert it to Angstroms, we multiply by \(10^{10}\): \[ \lambda = 1.735 \times 10^{10} \times 10^{-10} \mathrm{A} = 1.735\ \mathrm{ A} \] The answer is (B) \(1.735 \AA\).

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Most popular questions from this chapter

The mass of a particle is 400 times than that of an electron and charge is double. The particle is accelerated by \(5 \mathrm{~V}\). Initially the particle remained at rest, then its final kinetic energy is \(\ldots \ldots \ldots\) (A) \(5 \mathrm{eV}\) (B) \(10 \mathrm{eV}\) (C) \(100 \mathrm{eV}\) (D) \(2000 \mathrm{eV}\)

Find the velocity at which mass of a proton becomes \(1.1\) times its rest mass, \(\mathrm{m}_{\mathrm{g}}=1.6 \times 10^{-27} \mathrm{~kg}\) Also, calculate corresponding temperature. For simplicity, consider a proton as non- interacting ideal-gas particle at \(1 \mathrm{~atm}\) pressure. $\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \cdot \mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) $\mathrm{V}=1.28 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{12} \mathrm{~K}$ (B) $\mathrm{V}=12.6 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{11} \mathrm{~K}$ (C) $\mathrm{V}=1.26 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=5.76 \times 10^{11} \mathrm{~K}$ (D) $\mathrm{V}=12.6 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.56 \times 10^{11} \mathrm{~K}$

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) \(5.845 \times 10^{-12} \mathrm{~m}\)

When a radiation of wavelength \(3000 \AA\) is incident on metal, $1.85 \mathrm{~V}$ stopping potential is obtained. What will be threshed wave length of metal? $\left\\{\mathrm{h}=66 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(4539 \AA\) (B) \(3954 \AA\) (C) \(5439 \AA\) (D) \(4395 \AA\)

According to Rayleigh and Jeans the black body radiation in the cavity is system of (A) progressive electromagnetic waves (B) standing electromagnetic waves (C) electromagnetic waves of discrete (D) standing waves in lattice frequencies
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