Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The work function for caesium atom is 1.9 eV. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

2.51  The work function for caesium atom is 1.9 eV. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answers (1)

best_answer

Finding the kinetic energy of the ejected electrons:

K.E of the ejected photoelectron:

= h[\nu-\nu_{o}] = hc\left [\frac{1}{\lambda} -\frac{1}{\lambda_{o}} \right ]

=(6.626\times10^{-34}Js)(3.0\times10^8m/s)\left [ \frac{1}{500\times10^{-9}m} - \frac{1}{654\times10^{-9}m} \right ]

= 19.878\times10^{-26}Jm\left [ \frac{154}{327000}\times10^9 m^{-1}\right ]

= 9.361\times10^{-20}J

Finding the Velocity of the ejected electrons:

KE = \frac{1}{2}mv^2

Where, 

m = mass of electron

v = velocity of electron

Therefore, the velocity is given by,

v =\sqrt{ \frac{2KE}{m}}

v = \sqrt{\frac{2\times9.361\times10^{-20}J}{9.1\times10^{-31}kg}} = 4.52\times10^5\ m/s

Posted by

Divya Prakash Singh

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question