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13.Nuclei
medium
The rest mass of an electron as well as that of positron is $0.51\, MeV$. When an electron and positron are annihilate, they produce gamma-rays of wavelength(s)
A$0.012\, \mathring A$
B$0.024 \,\mathring A$
C$0.012 \,\mathring A\, to \, \infty$
D$0.024\, \mathring A \,to \,\infty$
Solution
(a) Since electron and positron annihilate
$\lambda = \frac{{hc}}{{{E_{Total}}}} $
$= \frac{{6.6 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}}{{(0.51 + 0.51) \times {{10}^6} \times 1.6 \times {{10}^{ – 19}}}}$
$ = 1.21 \times {10^{ – 12}}m = 0.012{\mathring A}$.
$\lambda = \frac{{hc}}{{{E_{Total}}}} $
$= \frac{{6.6 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}}{{(0.51 + 0.51) \times {{10}^6} \times 1.6 \times {{10}^{ – 19}}}}$
$ = 1.21 \times {10^{ – 12}}m = 0.012{\mathring A}$.
Standard 12
Physics
