Question

$\alpha$-particles accelerated by $3\times {10}^5$ V bombarded a boron target. This results in the following nuclear reaction:
${}_{2}H{e}^4+{}_5{B}^{10}\to {}_6{C}^{13}+{}_1{H}^1+\gamma$
If the combined energy of ${}^{13}Cand{H}_{1}is5\times {10}^{5}eV$, then the frequency of $\gamma$-rays is $x\times {10}^{5}eV.$ $1\times {10}^{5}eV$ energy is used in penetrating the nucleus.

Find the value of $x?$

$(He=4.0026amu,B=10.0129amu,C=13.0036amu,$ ${}_1{H}^1=1.008amu)$

Answer 1

Total mass before reaction $=4.0026+10.0129=14.0155amu$

Total mass after reaction $=13.0036+1.008=14.0116amu$

Mass decay during reaction $=14.0155-14.0116=0.0039amu$

Total energy given out $=0.0039\times 931MeV$
$=3.6309\times {10}^{6}eV$
$=3.6309\times {10}^6\times 1.3602\times {10}^{-19}J$
$=5.816\times {10}^{-13}J$

$E=hv$
$5.816\times {10}^{-13}=6.625\times {10}^{-34}\times v$

Frequency, $v=8.77\times {10}^{20}Hz$
$v=\frac{c}{\lambda }$
$\lambda =\frac{c}{v}=\frac{3.0\times {10}^8}{8.77\times {10}^{20}}=3.4\times {10}^{-13}m$

Energy supplied to $\alpha$-particle = $q\times v$

$=2\times 1.602\times {10}^{-19}\times 3\times {10}^5\times J$

$=\frac{2\times 1.602\times {10}^{-19}\times 3\times {10}^5}{1.602\times {10}^{-19}}eV$ $=6\times {10}^{5}eV$

This energy is used up to overpower the generation of nucleus and imparting energy of C and H atoms, i.e.,
$1\times {10}^{5}eV+5\times {10}^{5}eV=6\times {10}^{5}eV$
So, value of $x$ is $6.$