Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A γ-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident γ-ray. If E = B, show that this cannot happen. Hence, calculate how much bigger than B must be E be for such a process to happen.
Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A γ-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident γ-ray. If E = B, show that this cannot happen. Hence, calculate how much bigger than B must be E be for such a process to happen.
-
1 Answer
-
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- B.E= 2.2MeV
E-B=Kn+KP=
Conservation of momentum= Pn+PP= E/C
E=B
It only happen if Pn=Pp=0
Let E=B+X where X<
X= (E/C-PP)/2m+
2 + =0
Pp=
= 4E2/c2
Similar Questions for you
Q = [4 *4.0026 – 15.9994] *931.5 MeV
Q = 10.2 MeV
-(1)
for B,
for B,
-(2)
The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 × 8.5 + 120 × 8.5) - (240 × 7.6) MeV
Q = (2 × 120 × 8.5) - (240 × 7.6) MeV = 2040 - 1824 = 216 MeV.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers