Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2400

Wavelength \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\) are incident on two identical metal plates and photo electrons are emitted. If \(\lambda_{\mathrm{A}}=2 \lambda_{\mathrm{B}}\), the maximum kinetic energy of photo electrons is \(\ldots \ldots \ldots\) (A) \(2 \mathrm{~K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}\) (B) \(\mathrm{K}_{\mathrm{A}}<\left(\mathrm{K}_{\mathrm{B}} / 2\right)\) (C) \(\mathrm{K}_{\mathrm{A}}=2 \mathrm{~K}_{\mathrm{B}}\) (D) \(\mathrm{K}_{\mathrm{A}}>\left(\mathrm{K}_{\mathrm{B}} / 2\right)\)

Short Answer

Expert verified
(B) \(K_{A}< \left(K_{B} / 2\right)\)

Step by step solution

01

Recall the photoelectric effect equation

The photoelectric effect equation is given by: \[E_g = K_{max} + W\] where \(E_g\) is the energy of the incident photon, \(K_{max}\) is the maximum kinetic energy of the emitted photoelectron, and \(W\) is the work function of the metal plate (minimum energy required to release a photoelectron).
02

Recall the relationship between wavelength and energy

The energy of a photon is related to its wavelength (\(\lambda\)) by the Planck's equation: \[E_g = \frac{hc}{\lambda}\] where \(h\) is the Planck's constant and \(c\) is the speed of light.
03

Write the equations for the two photons

Using the above information, we can write the energy equations for the two incident lights with wavelengths \(\lambda_A\) and \(\lambda_B\): \[E_{gA} = \frac{hc}{\lambda_A} = K_{maxA} + W\] \[E_{gB} = \frac{hc}{\lambda_B} = K_{maxB} + W\]
04

Use the given relationship between the wavelengths

We are given that \(\lambda_A = 2\lambda_B\). Substitute this relationship in the equation for \(E_{gA}\): \[E_{gA} = \frac{hc}{2\lambda_B} = K_{maxA} + W\] Now solve for \(K_{maxA}\): \[K_{maxA} = \frac{hc}{2\lambda_B} - W\]
05

Compare the maximum kinetic energies

Let's compare \(K_{maxA}\) and \(K_{maxB}\): \[\frac{K_{maxA}}{K_{maxB}} = \frac{\frac{hc}{2\lambda_B} - W}{\frac{hc}{\lambda_B} - W}\] We want to find the condition that compares \(K_{maxA}\) and \(K_{maxB}\) to match one of the given options. Simplify the above expression: \[\frac{K_{maxA}}{K_{maxB}} = \frac{\frac{1}{2} - \frac{W\lambda_B}{hc}}{1 - \frac{W\lambda_B}{hc}}\]
06

Analyze the obtained expression

The fraction on the right side of the expression cannot exceed 1, as the work function (\(W\)) and the wavelength (\(\lambda_B\)) are both positive: \[\frac{W\lambda_B}{hc} \le 1\] Since \(K_{maxA}\) is multiplied by \(\frac{1}{2}\) on the numerator, the fraction on the left-hand side will always be less than 1: \[\frac{K_{maxA}}{K_{maxB}} < 1\] This means that the maximum kinetic energy of photoelectrons due to the incident light with wavelength \(\lambda_A\) is less than that due to the incident light with wavelength \(\lambda_B\): \[K_{maxA} < K_{maxB}\] Comparing our result with the given options, we can conclude that the correct answer is: (B) \(K_{A}< \left(K_{B} / 2\right)\)

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

Direction Read the following question choose if: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) Assertion is true but the Reason is false. (d) Both Assertion and Reason is false. Assertion : The de Broglie wavelength of an electron accelerated through 941 volts is \(0.4 \AA\). Reason: Higher the acceleration potentials of electron, smaller is the de Broglie wavelength. (A) a (B) \(\mathrm{b}\) (C) (D) \(\mathrm{d}\)

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (I) Energy of photon of wavelength \(\lambda\) is (P) \((\mathrm{E} / \mathrm{p})\) (II) The de Broglie wavelength associated (Q) \(\left(\mathrm{hf} / \mathrm{c}^{2}\right)\) with particle of momentum \(\mathrm{P}\) is (II) Mass of photon in motion is (R) (hc \(/ \lambda\) ) (IV) The velocity of photon of energy (S) \((\mathrm{h} / \mathrm{p})\) \(\mathrm{E}\) and momentum \(\mathrm{P}\) is (A) I - P, II - Q. III - R, IV - S (B) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$ (C) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{P}_{3} \mathrm{IV}-\mathrm{Q}$ (D) $\mathrm{I}-\mathrm{S}, \mathrm{II}-\mathrm{R}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$

de-Broglie wavelength of atom at T \(\mathrm{K}\) absolute temperature will be \(\ldots \ldots \ldots\) (A) \([\mathrm{h} /\\{\mathrm{mkT}\\}]\) (B) \([\mathrm{h} /\\{\sqrt{3} \mathrm{mKT}\\}]\) (C) \([\\{\sqrt{2} \mathrm{mKT}\\} / \mathrm{h}]\) (D) \(\sqrt{(2 \mathrm{mKT})}\)

Light of \(4560 \AA 1 \mathrm{~mW}\) is incident on photo-sensitive surface of \(\mathrm{Cs}\) (Cesium). If the quantum efficiency of the surface is \(0.5 \%\) what is the amount of photoelectric current produced? (A) \(1.84 \mathrm{~mA}\) (B) \(4.18 \mu \mathrm{A}\) (C) \(4.18 \mathrm{~mA}\) (D) \(1.84 \mu \mathrm{A}\)

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) \(1.534\)
See all solutions

Recommended explanations on English Textbooks

Creative Story

Read Explanation

Phonetics

Read Explanation

Essay Prompts

Read Explanation

Pragmatics

Read Explanation

Global English

Read Explanation

Linguistic Terms

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