Electons are emitted with the zero velocity from a metal surface when exposed to radiation of wavelength  6800 Å. Calculate threshold frequency (ν0) and work function (W0) of the metal.
 

Answers (1)

Given the wavelength of radiation is 6800\ \AA.

Energy\ given = Work\ function + Kinetic\ energy.

But the electrons are emitted with zero velocity from a metal surface when it is exposed to radiation. That means the kinetic energy will be zero.

So, the Threshold frequency \nu_{o} will be:

\nu_{o} = \frac{c}{\lambda_{o}} = \frac{3.0\times10^8 m/s}{6800\times10^{-10}m} = 4.14\times10^{14}s^{-1}

and the Work function will be:

W_{o} = h\nu_{o} = 6.626\times10^{-34}Js \times 4.14\times10^{14}s^{-1} = 2.92\times10^{-19}J

Preparation Products

Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions