Log out Go to App
Go to App

Learning Materials

Features

Discover

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).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electron is accelerated under a potential difference of \(64 \mathrm{~V}\), the de Broglie wave length associated with electron is $=\ldots \ldots \ldots \ldots . \AA$ (Use charge of election $=1.6 \times 10^{-19} \mathrm{C},=9.1 \times 10^{-31} \mathrm{Kg}$, mass of electrum \(\mathrm{h}=6.6623 \times 10^{-43} \mathrm{~J}\).sec \()\) (A) \(4.54\) (B) \(3.53\) (C) \(2.53\) (D) \(1.534\)

An electron moving with velocity \(0.6 \mathrm{c}\), then de-brogly wavelength associated with is \(\ldots \ldots \ldots\) (rest mars of electron, \(\mathrm{m}_{0}=9.1 \times 10^{-31}(\mathrm{k} / \mathrm{s})\) \(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) (A) \(3.24 \times 10^{-12} \mathrm{~m}\) (B) \(32.4 \times 10^{-12} \mathrm{~m}\) (C) \(320 \times 10^{-12} \mathrm{~m}\) (D) \(3.29 \times 10^{-14} \mathrm{~m}\)

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 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 \(2 \lambda\)

Work function of metal is \(2.5 \mathrm{eV}\) If wave length of light incident on metal plate is \(3000 \AA\), stopping potential of emitted electron will be....... $\left\\{\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(0.82 \mathrm{~V}\) (B) \(0.41 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(3.28 \mathrm{~V}\)
See all solutions

Recommended explanations on English Textbooks

Research and Composition

Read Explanation

Lexis and Semantics Summary

Read Explanation

Synthesis Essay

Read Explanation

English Language Study

Read Explanation

Email

Read Explanation

Language Acquisition

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free