Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2356

If alpha particle and deuteron move with velocity \(\mathrm{V}\) and $2 \mathrm{v}$, the ratio of their de-Brogle wavelength will be............ deuteron (A) \(1: \sqrt{2}\) (B) \(2: 1\) (C) \(1: 1\) (D) \(\sqrt{2}: 1\)

Short Answer

Expert verified
The ratio of the de Broglie wavelengths of the alpha particle and the deuteron is \(\sqrt{2}:1\).

Step by step solution

01

Find the de Broglie wavelength for the alpha particle

We know that de Broglie wavelength is given by \(\lambda=\frac{h}{mv}\). The mass of an alpha particle (which is a Helium nucleus) is \(4m_u\) (where \(m_u\) is the atomic mass unit), and we are given its velocity as V. So, we get the de Broglie wavelength of alpha particle as: \[\lambda_\alpha=\frac{h}{(4m_u)V}\]
02

Find the de Broglie wavelength for the deuteron

Similarly, for the deuteron (which is a nucleus consisting of one proton and one neutron), its mass will be \(2m_u\), and its given velocity is 2V. Therefore, the de Broglie wavelength of the deuteron would be: \[\lambda_d=\frac{h}{(2m_u)(2V)}\]
03

Calculate the ratio of wavelengths

Now, we need to find the ratio of the de Broglie wavelengths of the alpha particle and deuteron: \[\frac{\lambda_\alpha}{\lambda_d}=\frac{\frac{h}{(4m_u)V}}{\frac{h}{(2m_u)(2V)}}\] _multiplying both numerator and denominator by \((4m_u)V(2m_u)(2V)\)_ \[\frac{\lambda_\alpha}{\lambda_d}=\frac{h(2m_u)(2V)}{(4m_u)V}\]
04

Simplify the ratio

By cancelling the terms, we get: \[\frac{\lambda_\alpha}{\lambda_d}=\frac{2}{\sqrt{2}}\] This can be simplified: \[\frac{\lambda_\alpha}{\lambda_d}=\sqrt{2}:1\] So, the correct answer is (D) \(\sqrt{2}:1\).

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

Radius of a beam of radiation of wavelength \(5000 \AA\) is $10^{-3} \mathrm{~m}\(. Power of the beam is \)10^{-3} \mathrm{~W}$. This beam is normally incident on a metal of work function \(1.9 \mathrm{eV}\). The charge emitted by the metal per unit area in unit time is \(\ldots \ldots \ldots\) Assume that each incident photon emits one electron. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(1.282 \mathrm{C}\) (B) \(12.82 \mathrm{C}\) (C) \(128.2 \mathrm{C}\) (D) \(1282 \mathrm{C}\)

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) \(2000 \mathrm{eV}\)

Photoelectric effect is obtained on metal surface for a light having frequencies \(\mathrm{f}_{1} \& \mathrm{f}_{2}\) where \(\mathrm{f}_{1}>\mathrm{f}_{2}\). If ratio of maximum kinetic energy of emitted photo electrons is \(1: \mathrm{K}\), so threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) $\left\\{\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (B) $\left\\{\left(\mathrm{Kf}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (C) $\left\\{\left(\mathrm{K} \mathrm{f}_{2}-\mathrm{f}_{1}\right) /(\mathrm{K}-1)\right\\}$ (D) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)

An electron enters perpendicularly into uniform magnetic field having magnitude \(0.5 \times 10^{-5} \mathrm{~T}\). If it moves on a circular path of radius \(2 \mathrm{~mm}\), its de - Broglie wavelength is $\ldots \ldots . . . \AA$ (A) 3410 (B) 4140 (C) 2070 (D) 2785

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) \(5.75 \times 10^{8} \mathrm{MHz}\)
See all solutions

Recommended explanations on English Textbooks

Linguistic Terms

Read Explanation

Syntax

Read Explanation

Graphology

Read Explanation

English Grammar

Read Explanation

Morphology

Read Explanation

Rhetorical Analysis Essay

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