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: 12
        

Q.1.A beam of electrons accelerated with 4.64 V was passed through a tube having mercury vapours .As a result of absorption,electronic changes occured with mercury atoms and light was emitted.If the full energy of the single electron ws converted into light ,what was the wave number of emitted light.
Q.2.The binding energy of an electron in the ground state of an atom is equal to 24.6eV.Find the enegry required to remove both electrons from the atom.

10 years ago

Answers : (2)

askiitian.expert- chandra sekhar
10 Points
							

Hi Vinay,

1) 4.64 eV is the energy of the electron

    if its total energy is converted photons then

    hc/λ=4.64*1.602*10-19J

    1/λ=3.74*106


2)B.E is 24.6 eV

   for hydrogen it is 13.6 eV

   24.6 which is nearly twice that of hydrogen implies the atom is He

   for first electron it is 24.6 eV

   for second electron it is 27.2 eV

   for removing both the electrons it is 27.2+24.6=51.8 eV

 

10 years ago
Sankalp Samuel Das
33 Points
							

A.1)

V= 4.64V

E= QV

  = (1.6 x 10-19 x 4.64) J

  = 7.424 x 10-19 J

E=  hc /λ

  = hcv ,           [where v= 1/λ, v=wave number]

=> v= E/hc

=> v= (7.424 x 10-19) / (6.62 x 10-34 x 3 x 108) m-1

=> v= 37400 cm-1

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


Course Features

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


Course Features

  • 141 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