Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2410

The work function for tungsten and solidus are \(4.5 \mathrm{eV}\) and $2.3 \mathrm{eV}\( respectively. If the threshold wavelength \)\lambda_{0}$ for sodium is \(5460 \AA\) the value of \(\lambda_{0}\) for tungsten is $\ldots \ldots$ (A) \(528 \AA\) (B) \(2791 \AA\) (C) \(589 \AA\) (D) \(10683 \AA\)

Short Answer

Expert verified
The threshold wavelength \(\lambda_{0}\) for tungsten can be calculated using the given information. The work function ratio of tungsten and solidus is \(\frac{4.5}{2.3}\) and the threshold wavelength for sodium (solidus) is \(5460 \AA\). Using the relationship \(\frac{\lambda_{0_{solidus}}}{\lambda_{0_{tungsten}}} = \frac{W_{tungsten}}{W_{solidus}}\), we can solve for \(\lambda_{0_{tungsten}}\), which is equal to \(2791 \AA\). Therefore, the correct answer is \( \boxed{(B) 2791 \AA} \).

Step by step solution

01

(Step 1: Find the work function ratio of tungsten and solidus)

First, we will find the ratio of the work functions of tungsten and solidus: \[ \frac{W_{tungsten}}{W_{solidus}} = \frac{4.5}{2.3} \]
02

(Step 2: Calculate the ratio of threshold wavelengths)

Now, use the photoelectric effect equation for both tungsten and solidus. Divide the equation for tungsten by the equation for solidus. \[ \frac{hc/\lambda_{0_{tungsten}}}{hc/\lambda_{0_{solidus}}} = \frac{W_{tungsten}}{W_{solidus}} \] Since \(h\) and \(c\) are constants, they will cancel out, leaving: \[ \frac{\lambda_{0_{solidus}}}{\lambda_{0_{tungsten}}} = \frac{W_{tungsten}}{W_{solidus}} \]
03

(Step 3: Plug in the known values)

Now, plug in the known values for sodium's threshold wavelength (solidus) \(\lambda_{0_{solidus}} = 5460 \AA\) and the work function ratio calculated in step 1: \[ \frac{5460}{\lambda_{0_{tungsten}}} = \frac{4.5}{2.3} \]
04

(Step 4: Solve for the threshold wavelength of tungsten)

Now, we will isolate \(\lambda_{0_{tungsten}}\) by multiplying both sides of the equation by \(\lambda_{0_{tungsten}}\) and then dividing both sides by the work function ratio: \[ \lambda_{0_{tungsten}} = \frac{5460}{(4.5/2.3)} \] Solve the equation: \[ \lambda_{0_{tungsten}} = 2791 \AA \]
05

(Step 5: Choose the correct answer)

From the given options, the threshold wavelength of tungsten is \(2791 \AA\), which matches option (B). Therefore, the correct answer is (B) \(2791 \AA\).

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: Metals like Na or \(\mathrm{K}\), emit electrons even when visible lights fall on them. Reason: This is because their work function is low. (A) a (B) \(\mathrm{b}\) (C) \(\mathrm{c}\) (D) d

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

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) \(3.72\)

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) \(2000 \mathrm{eV}\)

An image of sun is formed by a lens of focal length \(30 \mathrm{~cm}\) on the metal surface of a photo-electric cell and a photoelectric current (I) is produced. The lens forming the image is then replaced by another of the same diameter but of focal length of \(15 \mathrm{~cm}\). The photoelectric current in this case is \(\ldots \ldots \ldots\) (A) \((1 / 2)\) (B) 1 (C) \(2 \mathrm{I}\) (D) \(4 \mathrm{I}\)
See all solutions

Recommended explanations on English Textbooks

Linguistic Terms

Read Explanation

Sociolinguistics

Read Explanation

Email

Read Explanation

Prosody

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

5 Paragraph 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