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
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !