Nuclear Physics And Radioactivity Question 51
Question: Let $ m_{p} $ be the mass of a proton, $ m_{n} $ the mass of a neutron, $ M_{1} $ the mass of a $ _{10}^{20}Ne $ nucleus and $ M_2 $ the mass of a $ _{20}^{40}Ca $ nucleus. Then [IIT 1998; DPMT 2000]
Options:
A) $ M_{2}=2M_{1} $
B) $ M_{2}>2M_{1} $
C) $ M_{2}<2M_{1} $
D) $ M_{1}<10(m_{n}+m_{p}) $
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Answer:
Correct Answer: C
Solution:
Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less then the sum of masses of it’s constituent particles $ _{10}^{20}Ne $ is made up of 10 protons plus 10 neutrons.
Therefore, mass of $ _{10}^{20}Ne $ nucleus $ M_1<10(m_p+m_n) $ Also heavier the nucleus,
more is he mass defect thus $ 20,(m_{n}+m_{p})-M_{2}>10(m_{p}+m_{n})-M_{1} $ or $ 10,(m_{p}+m_{n})>M_{2}-M_{1} $
Þ $ M_{2}<M_{1}+10,(m_{p}+m_{n}) $
Þ $ M_{2}<M_{1}+M_{1} $
Þ $ M_{2}<2M_{1} $ .
BETA
