MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: upto college level
        A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically upward component of velocity is non-positive for each photoelectron?
5 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
							Sol. When λ  = 250 nm
Energy of photon = hc/λ = 1240/250 = 4.96 ev
∴ K.E. – hc/λ = 1240/250 = 4.96 ev
Velocity to be non positive for each photo electron
The minimum value of velocity of plate should be = velocity of photo electron
∴ Velocity of photo electron = √2Ke/m
= √3.06/9.1 * 10^-31 = √3.06 * 1.6 * 10^-19/9.1 * 10^-31 = 1.04 * 10^6 m/sec.

						
5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details