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  • Given the masses of various atomic particles <img alt="m_{p}=1.0072\; u, m_{n}=1.0087 \; u,m_{e}=0.000548\; u, m_{v}^{-}=0,m_{d}=2.0141\; u," src="http://entrancecorner.codecogs.com/gif.latex?m_%7Bp%7D%3D1.0072%5C%3B%20u%2C%20m_%7Bn%7D%3D1.0087%20%5C%3

Given the masses of various atomic particles m_{p}=1.0072\; u, m_{n}=1.0087 \; u,m_{e}=0.000548\; u, m_{v}^{-}=0,m_{d}=2.0141\; u, where \\p\equiv proton,n\equiv neutron,e\equiv electron, \bar{v}\equiv antineutrino\; \; and\; \; d\equiv deuteron. Which of the following process is allowed by momentum and energy conservation ?
Option: 1 n+n\rightarrow deuterium atom (electron bound to the nucleus)
 
Option: 2 p\rightarrow n+e^{+}+\bar{v}
Option: 3 n+p\rightarrow d+\gamma  
Option: 4 e^{+}+e^{-}\rightarrow \gamma

Answers (1)

best_answer

\begin{aligned} &\Delta \mathrm{m} \text { should be positive }\\ &\left(m_{p}+m_{n}\right)>m_{d} \end{aligned}

\\ \text{Only in case-III,} \\\mathrm{M}_{\mathrm{LHS}}>\mathrm{M}_{\mathrm{RHS}}$ i.e. total mass on reactant side is greater then that on the product side. Hence it will only be allowed.

Posted by

Deependra Verma

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