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Chapter 17: Problem 2320

de-Broglie wavelength of proton accelerated under \(100 \mathrm{~V}\) electric potential difference is \(\lambda_{0} .\) If de-wave length \(\alpha\) -particle accelerated by the same electric potential difference what will its bouglie wavelength...... (A) \(2 \sqrt{2 \lambda_{0}}\) (B) \(\left\\{\lambda_{0} /(2 \sqrt{2})\right\\}\) (C) \(\left(\lambda_{0} / \sqrt{2}\right)\) (D) \(\left(\lambda_{0} / 2\right)\)

Short Answer

Expert verified
The short answer is: \(\lambda_\alpha = \frac{\lambda_0}{2\sqrt{2}}\), which corresponds to option (B).

Step by step solution

01

Write down the known values and the de Broglie wavelength formula

We know the de Broglie wavelength of the proton is given by \(\lambda_0\). The de Broglie wavelength formula is given by \(\lambda = \frac{h}{p}\), where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant, and \(p\) is the momentum of the particle.
02

Find the momentum of the proton

The momentum of a particle can be found by \(p = \sqrt{2mE}\), where \(m\) is the mass of the particle and \(E\) is its energy. The energy of a proton accelerated under a 100 V electric potential difference is given by \(E_p = eV\), where \(e\) is the elementary charge and \(V\) is the voltage. The mass of a proton is \(m_p\), so the momentum of the proton is \(p_p = \sqrt{2m_p(eV)}\).
03

Find the de Broglie wavelength of the proton

Now that we have the momentum of the proton, we can find its de Broglie wavelength using the formula \(\lambda = \frac{h}{p}\). We have \(\lambda_0 = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p(eV)}}\)
04

Find the momentum of the alpha particle

The mass of an alpha particle is \(m_\alpha = 4m_p\), since it consists of 2 protons and 2 neutrons, and each has a mass approximately equal to the mass of a proton. The energy of the alpha particle suspended under a 100 V electric potential difference is given by \(E_\alpha = 2eV\), since an alpha particle has a charge of \(+2e\). Therefore, the momentum of the alpha particle is given by \(p_\alpha = \sqrt{2m_\alpha(E_\alpha)} = \sqrt{2(4m_p)(2eV)}\).
05

Find the de Broglie wavelength of the alpha particle

Using the de Broglie formula, we have \(\lambda_\alpha = \frac{h}{p_\alpha} = \frac{h}{\sqrt{2(4m_p)(2eV)}}\).
06

Express the de Broglie wavelength of the alpha particle in terms of the de Broglie wavelength of the proton

Divide both wavelengths to obtain the relationship: \(\frac{\lambda_\alpha}{\lambda_0} = \frac{\frac{h}{\sqrt{2(4m_p)(2eV)}}}{\frac{h}{\sqrt{2m_p(eV)}}}\). Simplify the equation: \(\frac{\lambda_\alpha}{\lambda_0} = \frac{\sqrt{2m_p(eV)}}{\sqrt{2(4m_p)(2eV)}} = \frac{1}{2\sqrt{2}}\). Therefore, \(\lambda_\alpha = \frac{\lambda_0}{2\sqrt{2}}\), which corresponds to option (B).

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Most popular questions from this chapter

Work function of metal is \(2 \mathrm{eV}\). Light of intensity $10^{-5} \mathrm{Wm}^{-2}\( is incident on \)2 \mathrm{~cm}^{2}\( area of it. If \)10^{17}$ electrons of these metals absorb the light, in how much time does the photo electric effectc start? Consider the waveform of incident light. (A) \(1.4 \times 10^{7} \mathrm{sec}\) (B) \(1.5 \times 10^{7} \mathrm{sec}\) (C) \(1.6 \times 10^{7} \mathrm{sec}\) (D) \(1.7 \times 10^{7} \mathrm{sec}\)

If de-Broglie wavelength of electron is increased by \(1 \%\) its momentum \(\ldots \ldots\) (A) increases by \(1 \%\) (B) decreases by \(1 \%\) (C) increased by \(2 \%\) (D) decreases by \(2 \%\)

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) \(\sqrt{\left(m_{e} / m_{p}\right)}\)

With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is \(0.4 \AA\) (A) \(9410 \mathrm{~V}\) (B) \(94.10 \mathrm{~V}\) (C) \(9.140 \mathrm{~V}\) (D) \(941.0 \mathrm{~V}\)

At \(10^{\circ} \mathrm{C}\) temperature, de-Broglie wave length of atom is $0.4 \AA\(.If temperature of atom is increased by \)30^{\circ} \mathrm{C}$, what will be change in de-Broglie wavelength of atom? (A) decreases \(10^{-2} \AA\) (B) decreases \(2 \times 10^{-2} \AA\) (C) increases \(10^{-2} \AA\) (D) increases \(2 \times 10^{-2} \AA\)
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