Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2409

The work function of a metal is \(1 \mathrm{eV}\). Light of wavelength $3000 \AA$ is incident on this metal surface. The maximum velocity of emitted photoelectron will be \(\ldots \ldots .\) (A) \(10 \mathrm{~ms}^{-1}\) (B) \(10^{3} \mathrm{~ms}^{-1}\) (C) \(10^{4} \mathrm{~ms}^{-1}\) (D) \(10^{6} \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The maximum velocity of the emitted photoelectrons is approximately \(10^6 \text{~ms}^{-1}\) (option D) when light of wavelength \(3000 \mathring{A}\) is incident on a metal surface with a work function of \(1 \mathrm{eV}.\)

Step by step solution

01

Identify the given data

The work function of the metal (Φ) is given as 1 eV and the wavelength of the incident light (λ) as 3000 Å.
02

Convert given data into SI units and needed units

First, we need to convert the work function of the metal from electronvolt (eV) to Joules (J) and wavelength from Angstrom (Å) to meter (m). Φ = 1 eV × (1.6 × 10^{-19} J/eV) = 1.6 × 10^{-19} J λ = 3000 Å × (1 × 10^{-10} m/Å) = 3 × 10^{-7} m
03

Calculate the energy of the incident photons using Planck's equation

Planck's equation relates energy (E) with the wavelength (λ) of a photon using the formula: E = \(\frac{hc}{\lambda}\) where h is the Planck constant (6.63 × 10^{-34} J s) and c is the speed of light (3 × 10^8 m/s). E = \(\frac{(6.63 \times 10^{-34} \text{J s}) \times (3 \times 10^{8} \text{m/s})}{3 \times 10^{-7} \text{m}}\) E = 6.63 × 10^{-19} J
04

Calculate the maximum kinetic energy of the emitted photoelectrons

Using the photoelectric effect equation, we can find the maximum kinetic energy (K.E.) of the emitted photoelectrons: K.E. = E - Φ K.E. = 6.63 × 10^{-19} J - 1.6 × 10^{-19} J K.E. = 5.03 × 10^{-19} J
05

Calculate the maximum velocity of the photoelectrons

Using the kinetic energy equation, we can find the maximum velocity (v) of the emitted photoelectrons: K.E. = \(\frac{1}{2}mv^{2}\) where m is the mass of an electron (9.11 × 10^{-31} kg). Rearranging the equation to solve for v, we get: v = \( \sqrt{\frac{2 \times \text{K.E.}}{m}} \) v = \( \sqrt{\frac{2 \times 5.03 \times 10^{-19} \text{J}}{9.11 \times 10^{-31} \text{kg}}} \) v ≈ 10^6 m/s
06

Choose the correct answer from the given options

The maximum velocity of the emitted photoelectrons is approximately 10^6 m/s, which corresponds to option (D). Therefore, the correct answer is: (D) \(10^{6} \text{~ms}^{-1}\)

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 photon, an electron and a uranium nucleus all have same wavelength. The one with the most energy \(\ldots \ldots \ldots \ldots\) (A) is the photon (B) is the electron (C) is the uranium nucleus (D) depends upon the wavelength and properties of the particle.

De-Broglie wavelength of particle moving at a (1/4) th of speed of light having rest mass \(\mathrm{m}_{0}\) is \(\ldots \ldots \ldots\) (A) $\left\\{(3.87 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (B) $\left\\{(4.92 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (C) $\left\\{(7.57 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (D) $\left\\{(9.46 \mathrm{~h}) /\left(\mathrm{m}_{\circ} \mathrm{C}\right)\right\\}$

Power produced by a star is \(4 \times 10^{28} \mathrm{~W}\). If the average wavelength of the emitted radiations is considered to be \(4500 \AA\) the number of photons emitted in \(1 \mathrm{~s}\) is \(\ldots \ldots\) (A) \(1 \times 10^{45}\) (B) \(9 \times 10^{46}\) (C) \(8 \times 10^{45}\) (D) \(12 \times 10^{46}\)

A \(100 \mathrm{~W}\) bulb radiates energy at rate of $100 \mathrm{~J} / \mathrm{s}\(. If the light emitted has wavelength of \)525 \mathrm{~nm}$. How many photons are emitted per second? (A) \(1.51 \times 10^{20}\) (B) \(1.51 \times 10^{22}\) (C) \(2.64 \times 10^{20}\) (D) \(4.5 \times 10^{19}\)

In an experiment to determine photoelectric characteristics for a metal the intensity of radiation is kept constant. Starting with threshold frequency. Now, frequency of incident radiation is increased. It is observed that $\ldots \ldots \ldots$ (A) the number of photoelectrons increases (B) the energy of photoelectrons decreases (C) the number of photoelectrons decreases (D) the energy of photoelectrons increases.
See all solutions

Recommended explanations on English Textbooks

Language Analysis

Read Explanation

English Language Study

Read Explanation

English Grammar Summary

Read Explanation

Phonetics

Read Explanation

Syntax

Read Explanation

Language and Social Groups

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