Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2358

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) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)

Short Answer

Expert verified
The threshold frequency for the metal surface is \(f_0 = \left\\{\left(Kf_1 - f_2\right) / (K - 1)\right\\}\).

Step by step solution

01

Write down the given information

We are given the frequencies \(f_1\), \(f_2\) and the ratio of maximum kinetic energies \(1:K\). Let the maximum kinetic energy at frequency \(f_1\) be \(E_1\) and at frequency \(f_2\) be \(E_2\). Based on the given information: \[ \frac{E_1}{E_2} = \frac{1}{K} \]
02

Use the photoelectric effect formula

According to the photoelectric effect formula: For frequency \(f_1\), \[ E_1 = h(f_1 - f_0) \] For frequency \(f_2\), \[ E_2 = h(f_2 - f_0) \]
03

Plug energy ratio into the equations

We can plug the energy ratio into the equations: \[ \frac{E_1}{E_2} = \frac{h(f_1 - f_0)}{h(f_2 - f_0)} = \frac{1}{K} \]
04

Solve for the threshold frequency

Now we will solve the equation for the threshold frequency \(f_0\): \[ \frac{f_1 - f_0}{f_2 - f_0} = \frac{1}{K} \] Multiply both sides by \(K(f_2 - f_0)\): \[ K(f_1 - f_0) = f_2 - f_0 \] Now, solve for \(f_0\): \[ f_0 = \frac{Kf_1 - f_2}{K - 1} \] So the threshold frequency for the metal surface is: \[ f_0 = \left\\{\left(Kf_1 - f_2\right) / (K - 1)\right\\} \] The correct answer is (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

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) \(1.84 \mu \mathrm{A}\)

Particle \(\mathrm{A}\) and \(\mathrm{B}\) have electric charge \(+\mathrm{q}\) and \(+4 \mathrm{q} .\) Both have mass \(\mathrm{m}\). If both are allowed to fall under the same p.d., ratio of velocities $\left(\mathrm{V}_{\mathrm{A}} / \mathrm{V}_{\mathrm{B}}\right)=\ldots \ldots \ldots \ldots \ldots$ (A) \(2: 1\) (B) \(1: 2\) (C) \(1: 4\) (D) \(4: 1\)

An electric bulb of \(100 \mathrm{w}\) converts \(3 \%\) of electrical energy into light energy. If the wavelength of light emitted is \(6625 \AA\), the number of photons emitted is \(1 \mathrm{~s}\) is \(\ldots \ldots\) $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(10^{17}\) (B) \(10^{19}\) (C) \(10^{21}\) (D) \(10^{15}\)

A proton and electron are lying in a box having impenetrable walls, the ratio of uncertainty in their velocities are \(\ldots \ldots\) \(\left(\mathrm{m}_{\mathrm{e}}=\right.\) mass of electron and \(\mathrm{m}_{\mathrm{p}}=\) mass of proton. (A) \(\left(\mathrm{m}_{\mathrm{e}} / \mathrm{m}_{\mathrm{p}}\right)\) (B) \(\mathrm{m}_{\mathrm{e}} \cdot \mathrm{m}_{\mathrm{p}}\) (C) \(\left.\sqrt{\left(m_{e}\right.} \cdot m_{p}\right)\) (D) \(\sqrt{\left(m_{e} / m_{p}\right)}\)

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) \(\mathrm{K}_{\mathrm{A}}>\left(\mathrm{K}_{\mathrm{B}} / 2\right)\)
See all solutions

Recommended explanations on English Textbooks

Single Paragraph Essay

Read Explanation

English Grammar Summary

Read Explanation

Listening and Speaking

Read Explanation

Phonetics

Read Explanation

Linguistic Terms

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