badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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

                        

an electron is a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum required for its escape from the atom.what is the wavelength of emitted electron? A)4.70 ANGSTROM B)4.70 NANOMETER C)9.4 ANGSTROM D)9.40 NANOMETER

11 years ago

Answers : (7)

askiitian.expert- chandra sekhar
10 Points
							Hi Deepankar,

Energy required for the electron to escape is 13.6 eV
Excess energy that electron have is 0.5*13.6=6.8 eV
6.8 * 1.6 *10^-19 J = h c/wavelength
but the answers are not matching... there might may be some mistake... check it out
						
11 years ago
nishchith
30 Points
							
excess energy = 6.8 eV = 6.8 * 1.6 * 10^-19 J = 1.09 * 10^-18 J
h/mc = wavelength
here it can also be written as wavelength = h/sqrt(2mKE)
where h = 6.626 * 10^-34
m = 9.1 * 10^-31 kg
KE = eV = 1.09 * 10^-18 J
so substitute these answers and the right answer is 4.7 angstrom. 
 
5 years ago
nishchith
30 Points
							
excess energy = 6.8 eV = 6.8 * 1.6 * 10^-19 J = 1.09 * 10^-18 J
h/mc = wavelength
here it can also be written as wavelength = h/sqrt(2mKE)
where h = 6.626 * 10^-34
m = 9.1 * 10^-31 kg
KE = eV = 1.09 * 10^-18 J
so substitute these answers and the right answer is 4.7 angstrom. 
 
5 years ago
Heet Surana
26 Points
							Energy required for the electron to escape is 13.6 eV Excess energy that electron have is 0.5*13.6=6.8 eV 6.8 * 1.6 *10^-19 J = h c/wavelength but the answers are not matching... there might may be some mistake... check it out
						
4 years ago
Ankit
29 Points
							Excess energy = 13.6×.5 =6.8And lambda = square root of 150 / V  A°  where V is the potential difference supplied to electron So lambda = square. Rt 150/6.8 that gives lambda = 4.71 A°
						
2 years ago
Alankrita Singh
26 Points
							
 wavelength=12.27A°/√V 
So  V=13.6×1.5×1.6×10^-19
 Wavelength=4.7A° .........................................................
2 years ago
ankit singh
askIITians Faculty
596 Points
							
he energy given to electron is 13.6×1.5 = 20.4eV. So the kinetic energy of emitted electron is = 20.4-13.6 = 6.8eV. The wavelength of wave associated with electron moving with kinetic energy N eV is, (12.27) / (N^1/2) angstrom. 12.27/ (6.8)^1/2 = 4.7 angstrom.
4 months 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