Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2369

Ration of momentum of photons having wavelength \(4000 \AA \& 8000 \AA\) is ........... (A) \(2: 1\) (B) \(1: 2\) (C) \(20: 1\) (D) \(1: 20\)

Short Answer

Expert verified
The ratio of momenta of photons having wavelengths 4000 Å and 8000 Å is 2:1. The correct answer is (A) \(2: 1\).

Step by step solution

01

Convert Wavelengths

First, let's convert the given wavelengths from Angstrom to meters. 1 Å = \(10^{-10}\) meters So the wavelengths in meters are: λ₁ = 4000 Å = 4000 × \(10^{-10}\) m = \(4 \times 10^{-7}\) m λ₂ = 8000 Å = 8000 × \(10^{-10}\) m = \(8 \times 10^{-7}\) m
02

Calculate Momentum

Now, let's find the momentum of the two photons using the formula: momentum (p) = \(\frac{h}{λ}\) where h is the Planck constant, h = \(6.63 \times 10^{-34}\) Js For Photon 1 with λ₁: \(p_1 = \frac{6.63 \times 10^{-34}}{4 \times 10^{-7}}\) For Photon 2 with λ₂: \(p_2 = \frac{6.63 \times 10^{-34}}{8 \times 10^{-7}}\)
03

Find the Ratio

Now, we will find the ratio of the momenta, \(\frac{p_1}{p_2} = \frac{\frac{6.63 \times 10^{-34}}{4 \times 10^{-7}}}{\frac{6.63 \times 10^{-34}}{8 \times 10^{-7}}}\) Notice that the Planck constant (h) on both numerator and denominator cancels out: \(\frac{p_1}{p_2} = \frac{\frac{1}{4 \times 10^{-7}}}{\frac{1}{8 \times 10^{-7}}}\) Now, we can simplify the ratio by multiplying both numerator and denominator by \(8 \times 10^{-7}\): \(\frac{p_1}{p_2} = \frac{2}{1}\)
04

Final Answer

Thus, the ratio of momenta of photons having wavelengths 4000 Å and 8000 Å is 2:1. The correct answer is (A) \(2: 1\).

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

In photo electric effect, if threshold wave length of a metal is \(5000 \AA\) work function of this metal is ..........V. $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(1.24\) (B) \(2.48\) (C) \(4.96\) (D) \(3.72\)

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

Energy corresponding to threshed frequency of metal is \(6.2 \mathrm{eV}\). If stopping potential corresponding to radiation incident on surface is $5 \mathrm{~V}\(, incident radiation will be in the \)\ldots \ldots \ldots \ldots \ldots$ region. (A) X-ray (B) Ultraviolet (C) infrared (D) Visible

In photoelectric effect, work function of martial is \(3.5 \mathrm{eV}\). By applying \(-1.2 \mathrm{~V}\) potential, photo electric current becomes zero, so......... (A) energy of incident photon is \(4.7 \mathrm{eV}\). (B) energy of incident photon is \(2.3 \mathrm{eV}\) (C) If photon having higher frequency is used, photo electric current is produced. (D) When energy of photon is \(2.3 \mathrm{eV}\), photo electric current becomes maximum

Work function of metal is \(4.2 \mathrm{eV}\) If ultraviolet radiation (photon) having energy \(6.2 \mathrm{eV}\), stopping potential will be........ (A) \(2 \mathrm{eV}\) (B) \(2 \mathrm{~V}\) (C) 0 (d) \(10.4 \mathrm{~V}\)
See all solutions

Recommended explanations on English Textbooks

Summary Text

Read Explanation

History of English Language

Read Explanation

Single Paragraph Essay

Read Explanation

Research and Composition

Read Explanation

Creative Story

Read Explanation

Rhetoric

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