IIT JAM
PreviousNext
Vijay Dalakoti Asked a Question
January 11, 2022 7:17 pmpts 30 pts
  • IIT JAM
  • Physics (PH)

Please prove the solution part? how did that expression came? en=v0+n2h2/8ml2?

A particle of mass m and energy E is incident on the following 1-D potential 0 for x<0 V(x)={V for 0 L The minimum value of E for which reflection coefficient will be zero, is h2 V+ mL O O V+ 2mL O h2 V+ 4mL O h V+ 8mL The reflection co-eficient becomes zero for energies of the form: nh E, = Vo + 8m? =,2, 3,... h Therefore, the minmm value is. E = Vo+ SmL
  • 1 Answer(s)
Answer Now
  • 0 Likes
  • 1 Comments
  • Shares
  • Bidyut sinha thankyou
    please see attached picture.
    • cropped1135112557.jpg
    Likes(0) Reply(0)
How Eduncle Helps You ?
StudentTutorInstitutionApp
Help
FAQsPayment TermsRefund PolicyAsk Support
Head Office :
MPA 44, 2nd floor, Rangbari Main Road,
Mahaveer Nagar II, Kota (Raj.) – 324005

Corporate Office:
212, F-1, 2nd Floor, Evershine Tower,
Amrapali Marg,
Vaishali Nagar, Jaipur (Raj.) – 302021

Mail: info@eduncle.com
All Rights Reserved © Eduncle.com