Given the masses of various atomic particles m? = 1.0072u, m? = 1.0087u, m? = 0.000548u, m? = 0, m? = 2.0141u, where p = proton, n = neutron, e = electron, v = antineutrino and d = deuteron. Which of the following process is allowed by momentum and energy conservation?
Given the masses of various atomic particles m? = 1.0072u, m? = 1.0087u, m? = 0.000548u, m? = 0, m? = 2.0141u, where p = proton, n = neutron, e = electron, v = antineutrino and d = deuteron. Which of the following process is allowed by momentum and energy conservation?
Option 1 -
n + n → deuterium atom (electron bound to the nucleus)
Option 2 -
p → n + e⁺ + v
Option 3 -
n + p → d + γ
Option 4 -
e⁺ + e⁻ → γ
Asked by Shiksha User
-
1 Answer
-
Correct Option - 3
Detailed Solution:Total mass of reactant should be greater than that of product.
This condition is only fulfilled in case- 3
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