Question

Given the masses of various atomic particles ${\mathrm{m}}_{\mathrm{p}}=1.0072\mathrm{u},{\mathrm{m}}_{\mathrm{n}}=1.0087\mathrm{u},{\mathrm{m}}_{\mathrm{e}}=0.000548\mathrm{u},{\mathrm{m}}_{\mathrm{v}\left(\mathrm{bar}\right)}=0,{\mathrm{m}}_{\mathrm{d}}=2.0141\mathrm{u},$where $\mathrm{p}\equiv \mathrm{proton},\mathrm{n}\equiv \mathrm{neutron},\mathrm{e}\equiv \mathrm{electron},\mathrm{v}\left[\mathrm{bar}\right]\equiv \mathrm{antineutrino}$ and $\mathrm{d}\equiv \mathrm{deuteron}.$ Which of the following processes is allowed by momentum and energy conservation?

• A. $\mathrm{n}+\mathrm{n}\mathrm{deuterium}\mathrm{atom}$ (electron bound to the nucleus)
• B. ${\mathrm{e}}^{+}+{\mathrm{e}}^{-}\to \mathrm{\gamma }$
• C. $\mathrm{p}\to \mathrm{n}+{\mathrm{e}}^{+}+\mathrm{v}\left[\mathrm{bar}\right]$
• D. $\mathrm{n}+\mathrm{p}\to \mathrm{d}+\mathrm{\gamma }$

Answer 1

The correct option is D

$\mathrm{n}+\mathrm{p}\to \mathrm{d}+\mathrm{\gamma }$

Answer(D)

Explanation for correct option:

(D) For the momentum and energy conservation, the mass defect $(∆\mathrm{m})$ should be positive. Since some energy is lost in every process.

Here option (D) has a positive mass defect because $({\mathrm{m}}_{\mathrm{p}}+{\mathrm{m}}_{\mathrm{m}})>{\mathrm{m}}_{\mathrm{d}}$

Explanation for incorrect options:

(A) Incorrect option because $\mathrm{n}+\mathrm{p}\to \mathrm{d}$

(B) Incorrect option because ${\mathrm{e}}^{-}+{\mathrm{e}}^{-}\to \mathrm{\gamma }$

(C) Incorrect option because $\mathrm{mass}\uparrow$

Hence, option D is the correct option.