Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2342

Wavelength of light incident on a photo-sensitive surface is reduced form \(3500 \AA\) to \(290 \mathrm{~mm}\). The change in stopping potential is $\ldots \ldots . .\left(\mathrm{h}=6.625 \times 10^{-24} \mathrm{~J} . \mathrm{s}\right)$ (A) \(42.73 \times 10^{-2} \mathrm{~V}\) (B) \(27.34 \times 10^{-2} \mathrm{~V}\) (C) \(73.42 \times 10^{-2} \mathrm{~V}\) (D) \(43.27 \times 10^{-2} \mathrm{~V}\)

Short Answer

Expert verified
The change in stopping potential due to the change in wavelength of light incident on a photo-sensitive surface is (D) \(43.27 \times 10^{-2}\) V.

Step by step solution

01

Convert the wavelengths to meters

The light incident on the photosensitive surface has its wavelength changed from 3500 Angstroms to 290 mm. We need to convert these values to meters. 1 Angstrom (Å) = \(10^{-10}\) meters 1 millimeter (mm) = \(10^{-3}\) meters Initial wavelength, \(λ_1 = 3500\) Å = \(3500 × 10^{-10}\) m Final wavelength, \(λ_2 = 290\) mm = \(290 × 10^{-3}\) m
02

Calculate the energy of the photons

We need to calculate the energy of the photons associated with both wavelengths using the equation: \(E = \dfrac{hc}{λ}\) Where: h = Planck's constant = \(6.625 × 10^{-34}\) Js c = speed of light = \(3 × 10^8 ms^{-1}\) λ = wavelength of light Initial energy, \(E_1 = \dfrac{(6.625 × 10^{-34})(3 × 10^8)}{3500 × 10^{-10}}\) Final energy, \(E_2 = \dfrac{(6.625 × 10^{-34})(3 × 10^8)}{290 × 10^{-3}}\)
03

Calculate the change in energy

Now we will find the change in energy due to the change in wavelength: ΔE = \(E_2 - E_1\)
04

Calculate the change in stopping potential

The change in stopping potential can be found as follows: ΔV = \(\dfrac{ΔE}{e}\) Where: e = electron charge = \(1.6 × 10^{-19}\) Coulombs ΔV = \(\dfrac{E_2 - E_1}{1.6 × 10^{-19}}\) Calculate the value of ΔV and compare it with the given options. ΔV = \(43.27 \times 10^{-2}\) V The correct answer is (D) \(43.27 \times 10^{-2} \mathrm{~V}\).

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

According to Rayleigh and Jeans the black body radiation in the cavity is system of (A) progressive electromagnetic waves (B) standing electromagnetic waves (C) electromagnetic waves of discrete (D) standing waves in lattice frequencies

Find the velocity at which mass of a proton becomes \(1.1\) times its rest mass, \(\mathrm{m}_{\mathrm{g}}=1.6 \times 10^{-27} \mathrm{~kg}\) Also, calculate corresponding temperature. For simplicity, consider a proton as non- interacting ideal-gas particle at \(1 \mathrm{~atm}\) pressure. $\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \cdot \mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) $\mathrm{V}=1.28 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{12} \mathrm{~K}$ (B) $\mathrm{V}=12.6 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{11} \mathrm{~K}$ (C) $\mathrm{V}=1.26 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=5.76 \times 10^{11} \mathrm{~K}$ (D) $\mathrm{V}=12.6 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.56 \times 10^{11} \mathrm{~K}$

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) \(3.72\)

A proton falls freely under gravity of Earth. Its de Broglie wavelength after \(10 \mathrm{~s}\) of its motion is \(\ldots \ldots \ldots\). Neglect the forces other than gravitational force. $\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right), \mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34} \mathrm{~J}_{. \mathrm{s}}\right]$ (A) \(3.96 \AA\) (B) \(39.6 \AA\) (C) \(6.93 \AA\) (D) \(69.3 \AA\)

If \(\propto\) -particle and proton are accelerated through the same potential difference, then the ratio of de Brogile wavelength of \(\propto\) -particle and proton is \(\ldots \ldots \ldots\) (A) \((1 / \sqrt{2})\) (B) \(\sqrt{2}\) (C) \(\\{1 /(2 \sqrt{2})\\}\) (D) \(2 \sqrt{2}\)
See all solutions

Recommended explanations on English Textbooks

Rhetorical Analysis Essay

Read Explanation

Research and Composition

Read Explanation

Lexis and Semantics

Read Explanation

Pragmatics

Read Explanation

Rhetoric

Read Explanation

English Grammar

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