Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2326

Photocell cell is enlightended by small bright source \(1 \mathrm{~m}\) away. If the same light source is placed \((1 / 2) \mathrm{m}\) away, number of electrons emitted by cathode will be........ (A) increases twice (B) decreases twice (C) increases 4 times (D) decreases 4 times

Short Answer

Expert verified
The number of electrons emitted by the cathode will increase 4 times when the same light source is placed at half the distance from the photocell. This is because the new intensity of light becomes 4 times stronger due to the inverse square law, which directly impacts the number of emitted electrons. The correct answer is (C) increases 4 times.

Step by step solution

01

Understand the inverse square law for light intensity

The inverse square law states that the intensity (I) of light is proportional to the inverse square of the distance (d) from the light source. Mathematically, it is given by: \(I ∝ \frac{1}{d^2}\) When the distance changes, the intensity of light also changes. We can use this principle to find the new intensity of light (I') when the light source is placed at half the distance.
02

Determine the change in the intensity of light

Initially, the light source is placed at a distance of 1 m. The new distance d' is given as: \(d'= \frac{1}{2} m \) To determine the change in intensity, first find the ratio between the initial and new distance squares: \(\frac{d'^2}{d^2} = \frac{\left(\frac{1}{2}\right)^2}{1^2}\) Now, simplify the expression: \(\frac{d'^2}{d^2} = \frac{1}{4}\) According to the inverse square law, the intensity of light is inversely proportional to the square of the distance. Therefore, the new intensity (I') will be 4 times stronger than the initial intensity (I).
03

Calculate the change in the number of emitted electrons

The number of emitted electrons is directly proportional to the intensity of light. As the new intensity is 4 times stronger than the initial intensity, the number of emitted electrons will also increase by 4 times. So, the correct answer is (C) increases 4 times.

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

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) \(\sqrt{\left(m_{e} / m_{p}\right)}\)

If we take accelerating voltage \(\mathrm{V}=50 \mathrm{~V}\), electric charge of electron \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\) and mass of electron \(\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}\) find the wavelength of concerned election. (A) \(0.1735 \mathrm{~A}\) (B) \(1.735 \AA\) (C) \(17.35 \AA\) (D) \(1735 \AA\)

Which of the following phenomenon can not be explained by quantum theory of light? (A) Emission of radiation from black body (B) Photo electric effect (C) Polarization (D) Crompton effect

U. V. light of wavelength \(200 \mathrm{~nm}\) is incident on polished surface of Fe. work function of the surface is \(4.5 \mathrm{eV}\). Find maximum speed of phote electrons. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(7.75 \times 10^{4}(\mathrm{~m} / \mathrm{s})\) (B) \(875 \times 10^{5}(\mathrm{~m} / \mathrm{s})\) (C) \(8.75 \times 10^{4}(\mathrm{~m} / \mathrm{s})\) (D) \(7.75 \times 10^{5}(\mathrm{~m} / \mathrm{s})\)

A star which can be seen with naked eye from Earth has intensity $1.6 \times 10^{-9} \mathrm{Wm}^{-2}\( on Earth. If the corresponding wavelength is \)560 \mathrm{~nm}\(, and the radius of the human eye is \)2.5 \times 10^{-3} \mathrm{~m}\(, the number of photons entering in our in \)1 \mathrm{~s}$ is..... (A) \(7.8 \times 10^{4} \mathrm{~s}^{-1}\) (B) \(8.85 \times 10^{4} \mathrm{~s}^{-1}\) (C) \(7.85 \times 10^{5} \mathrm{~s}^{-1}\) (D) \(8.85 \times 10^{5} \mathrm{~s}^{-1}\)
See all solutions

Recommended explanations on English Textbooks

Blog

Read Explanation

Language and Social Groups

Read Explanation

Textual Analysis

Read Explanation

Syntax

Read Explanation

Research and Composition

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