Given the masses of various atomic particles Mp=1.0072

Question:

Given the masses of various atomic particles $m_{p}=1.0072$

$\mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.0087 \mathrm{u}, \mathrm{m}_{\mathrm{e}}=0.000548 \mathrm{u}, \mathrm{m}_{v}^{-}=0, \mathrm{~m}_{\mathrm{d}}=2.0141 \mathrm{u}$,

where $\mathrm{p} \equiv$ proton, $\mathrm{n} \equiv$ neutron, $\mathrm{e} \equiv$ electron,

$\bar{v} \equiv$ antineutrino and $\mathrm{d} \equiv$ deuteron. Which of the following

process is allowed by momentum and energy conservation?

  1. (1) $n+n \rightarrow$ deuterium atom (electron bound to the nucleus)

  2. (2) $\mathrm{p} \rightarrow \mathrm{n}+\mathrm{e}^{+}+v$

  3. (3) $\mathrm{n}+\mathrm{p} \rightarrow \mathrm{d}+\gamma$

  4. (4) $\mathrm{e}^{+}+\mathrm{e}^{-} \rightarrow \gamma$


Correct Option: , 3

Solution:

(3) For the momentum and energy conservation, mass defect $(\Delta m)$ should be positive. Since some energy is lost in every process.

$\left(m_{p}+m_{n}\right)>m_{d}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now