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Chapter 17: Problem 2319

If velocity of free electron is made double, change in its de-Broglie wavelength will be \(\ldots \ldots .\) (A) increase by \((\lambda / 2)\) (B) decrease by \((\lambda 2)\) (C) increase by \(2 \lambda\) (D) decrease \(2 \lambda\)

Short Answer

Expert verified
The de-Broglie wavelength of the electron decreases by \((\frac{\lambda}{2})\).

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 de-Broglie wavelength - \(h\) is the Planck's constant and - \(p\) is the momentum of particle. Since \(p=mv\), where \(m\) is the mass and \(v\) is the velocity, the formula becomes: \[\lambda = \frac{h}{mv}\]
02

Calculate the initial de-Broglie Wavelength of the electron

Let the initial velocity of the electron be denoted by \(v\), then the initial de-Broglie wavelength \(\lambda_i\) is given by: \[\lambda_i = \frac{h}{mv}\]
03

Calculate the final de-Broglie Wavelength of the electron

Now, let's double the velocity of the electron. So, the final velocity of the electron is \(2v\). The final de-Broglie wavelength \(\lambda_f\) is given by: \[\lambda_f = \frac{h}{m(2v)}\]
04

Compare the initial and final de-Broglie Wavelengths to determine the change

Now that we have expressions for the initial and final wavelengths, we can compare them to determine the change. We can simplify the ratio of the initial and final wavelengths as follows: \[\frac{\lambda_f}{\lambda_i} = \frac{\frac{h}{m(2v)}}{\frac{h}{mv}} = \frac{1}{2}\] This indicates that the final wavelength is half the initial wavelength.
05

Find the change in de-Broglie Wavelength

Now that we found the ratio of the initial and final wavelengths, we can find the change in the wavelength. Since \(\lambda_f\) is \(\frac{1}{2}\) of \(\lambda_i\), the change in wavelength is: \[\Delta \lambda = \lambda_f - \lambda_i = -\frac{1}{2}\lambda_i\] Therefore, the de-Broglie wavelength of the electron decreases by \((\frac{\lambda}{2})\). The correct answer is: (A) increase by \((\lambda / 2)\)

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Most popular questions from this chapter

Photoelectric effect on surface is found for frequencies $5.5 \times 10^{8} \mathrm{MHz}\( and \)4.5 \times 10^{8} \mathrm{MHz}$ If ratio of maximum kinetic energies of emitted photo electrons is \(1: 5\), threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) \(7.55 \times 10^{8} \mathrm{MHz}\) (B) \(4.57 \times 10^{8} \mathrm{MHz}\) (C) \(9.35 \times 10^{8} \mathrm{MHz}\) (D) \(5.75 \times 10^{8} \mathrm{MHz}\)

If kinetic energy of free electron is made double, change in de-Broglie wavelength will be........... (A) \(\sqrt{2}\) (B) \((1 / \sqrt{2})\) (C) 2 (D) \((1 / 2)\)

When electric bulb having \(100 \mathrm{~W}\) efficiency emits photon having wavelength \(540 \mathrm{~nm}\) every second, numbers of photons will be $\ldots \ldots\left(\mathrm{h}=6 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) 100 (B) 1000 (C) \(3 \times 10^{20}\) (D) \(3 \times 10^{18}\)

Light of \(4560 \AA 1 \mathrm{~mW}\) is incident on photo-sensitive surface of \(\mathrm{Cs}\) (Cesium). If the quantum efficiency of the surface is \(0.5 \%\) what is the amount of photoelectric current produced? (A) \(1.84 \mathrm{~mA}\) (B) \(4.18 \mu \mathrm{A}\) (C) \(4.18 \mathrm{~mA}\) (D) \(1.84 \mu \mathrm{A}\)

Output power of He-Ne LASER of low energy is \(1.00 \mathrm{~mW}\). Wavelength of the light is \(632.8 \mathrm{~nm}\). What will be the number of photons emitted per second from this LASER? (A) \(8.31 \times 10^{15} \mathrm{~s}^{-1}\) (B) \(5.38 \times 10^{15} \mathrm{~s}^{-1}\) (C) \(1.83 \times 10^{15} \mathrm{~s}^{-1}\) (D) \(3.18 \times 10^{15} \mathrm{~s}^{-1}\)
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