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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

In which of the following phenomena the photon picture is required? (A) Energy distribution in black body radiation (B) Compton scattering (C) Photoelectric effect (D) all of the above

A proton and electron are lying in a box having impenetrable walls, the ratio of uncertainty in their velocities are \(\ldots \ldots\) \(\left(\mathrm{m}_{\mathrm{e}}=\right.\) mass of electron and \(\mathrm{m}_{\mathrm{p}}=\) mass of proton. (A) \(\left(\mathrm{m}_{\mathrm{e}} / \mathrm{m}_{\mathrm{p}}\right)\) (B) \(\mathrm{m}_{\mathrm{e}} \cdot \mathrm{m}_{\mathrm{p}}\) (C) \(\left.\sqrt{\left(m_{e}\right.} \cdot m_{p}\right)\) (D) \(\sqrt{\left(m_{e} / m_{p}\right)}\)

In an experiment to determine photoelectric characteristics for a metal the intensity of radiation is kept constant. Starting with threshold frequency. Now, frequency of incident radiation is increased. It is observed that $\ldots \ldots \ldots$ (A) the number of photoelectrons increases (B) the energy of photoelectrons decreases (C) the number of photoelectrons decreases (D) the energy of photoelectrons increases.

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