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Chapter 17: Problem 2399

Wavelength of an electron having energy \(10 \mathrm{ke} \mathrm{V}\) is $\ldots \ldots . \AA$ (A) \(0.12\) (B) \(1.2\) (C) 12 (D) 120

Short Answer

Expert verified
The wavelength of an electron having an energy of 10 keV is approximately 1.2 Å, which corresponds to option (B).

Step by step solution

01

Convert energy to Joules

First, we have to convert the given electron energy from keV to Joules. We know that 1 keV = \(1.6 \times 10^{-16}\) Joules. So, the electron energy E in Joules can be given as: E = 10 keV × \(1.6 \times 10^{-16}\) J/keV
02

Find the electron's momentum

We know that E = mc^2. We can rearrange the equation to find the momentum of the electron p: p = \(\frac{E}{c}\), where c is the speed of light (\(3 \times 10^8\) m/s)
03

Calculate the de Broglie wavelength

The de Broglie wavelength formula is: λ = \(\frac{h}{p}\), where λ is the wavelength, h is Planck's constant (\(6.626 \times 10^{-34}\) Js), and p is the electron's momentum We can now substitute the known values to find the wavelength of the electron: λ = \(\frac{h}{\frac{E}{c}}\) = \(\frac{hc}{E}\)
04

Express the wavelength in Angstroms

The wavelength λ in Angstroms (Å) can be found by converting the calculated wavelength from meters to Angstroms, using the relation 1 Å = \(1 \times 10^{-10}\) meters. So, λ in Å is: λ (Å) = λ (m) × \(\frac{1 \times 10^{10}\) Å}{1 m}\) Now, plug in the calculated values and find the wavelength in Angstroms: λ (Å) = \(\frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{ 10 \cdot 1.6 \times 10^{-16}} \times 1 \times 10^{10}\) λ (Å) ≈ 1.2 Å Thus, the wavelength of an electron having an energy of 10 keV is approximately 1.2 Å, which corresponds to option (B).

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Most popular questions from this chapter

Energy of photon having wavelength \(\lambda\) is \(2 \mathrm{eV}\). This photon when incident on metal. Maximum velocity of emitted is \(\mathrm{V}\). If \(\lambda\) is decreased \(25 \%\) and maximum velocity is made double, work function of metal is \(\ldots \ldots \ldots . . \mathrm{V}\) (A) \(1.2\) (B) \(1.5\) (C) \(1.6\) (D) \(1.8\)

de-Broglie wavelength of atom at T \(\mathrm{K}\) absolute temperature will be \(\ldots \ldots \ldots\) (A) \([\mathrm{h} /\\{\mathrm{mkT}\\}]\) (B) \([\mathrm{h} /\\{\sqrt{3} \mathrm{mKT}\\}]\) (C) \([\\{\sqrt{2} \mathrm{mKT}\\} / \mathrm{h}]\) (D) \(\sqrt{(2 \mathrm{mKT})}\)

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) $\mathrm{V}=12.6 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.56 \times 10^{11} \mathrm{~K}$

How many photons of red colored light having wavelength \(8000 \AA\) will have same energy as one photon of violet colored light of wavelength \(4000 \AA\) ? (A) 2 (B) 4 (C) 6 (D) 8

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) \(2000 \mathrm{eV}\)
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