JEE PYQ: Nuclear Physics Question 13
Question 13 - 2020 (04 Sep Shift 2)
Find the Binding energy per nucleon for $^{120}{50}\text{Sn}$. Mass of proton $m_p = 1.00783$ U, mass of neutron $m_n = 1.00867$ U and mass of tin nucleus $m{\text{Sn}} = 119.902199$ U. (take 1U = 931 MeV)
(1) 7.5 MeV
(2) 9.0 MeV
(3) 8.0 MeV
(4) 8.5 MeV
Show Answer
Answer: (4)
Solution
Mass defect, $\Delta m = (50 m_p + 70 m_n) - (m_{\text{Sn}})$ $= (50 \times 1.00783 + 70 \times 1.008) - (119.902199)$ $= 1.096$ Binding energy $= (\Delta m)c^2 = (\Delta m) \times 931 = 1020.56$ $\frac{\text{Binding energy}}{\text{Nucleon}} = \frac{1020.5631}{120} = 8.5$ MeV
BETA
