Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
A stationary nucleus of mass 24u emits a ¥ photon. The energy of the emited photon is 7MeV. Find the recoil energy of the nucleus?(ans:-1.1MeV).
in gama decay there is no change in mass number...
initially nucleous was at rest ,due to internal forces it will move after giving a gamma photon..
so we can apply conservation of momentam here
final momentam = initial momentam = 0
P(gamma photon) + P(nucleous) = 0 .......................1
energy of photon = 7Mev
p(photon) = E/c (c is speed of light , E =7Mev (given) )
p(nucleous) = -E/c ( from eq 1)
kinetic energy = P2/2m = E2/2mc2
now put E , C & m ,u will get the 1.1Mev ans
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !