Question

$\alpha$-particle accelerated by $3\times {10}^5$ volt bombarded a boron target. This results in the nuclear reaction.
${}_2^4\text{He}{+}_5^{10}\text{B}{⟶}_6^{13}\text{C}{+}_1^1\text{H}+\gamma$. If the combined energy of ${}^{13}\text{C}$ and ${}^1\text{H}$ is $5\times {10}^5$ eV, calculate the value of $\lambda$ of $\gamma$-rays (expressed as $\times {10}^{-14}m$) . $1\times {10}^5$eV energy is used in penetrating the nucleus.

[Given atomic weight of H = 1.008 amu, He = 4.0026 amu, B = 10.0129 amu, C = 13.0036 amu and 1 amu = 931 MeV.]

Answer 1

Energy supplied to $\alpha$-particle $q\times V$
$=2\times 1.602\times {10}^{-19}\times 3\times {10}^5$J
$=\frac{2\times 1.602\times {10}^{-19}\times 3\times {10}^5}{1.602\times {10}^{-19}}eV=6\times {10}^{5}eV$

The energy given is used up to overpower the penetration of nucleus and imparting energy to C and H atoms, i.e., $1\times {10}^{5}eV+5\times {10}^{5}eV=6\times {10}^{5}eV$.

Thus, extra energy given to $\alpha$-particles is used in imparting velocity to C and H and to overpower the forces of repulsion. The mass decayed during the course of the reaction is responsible for emission $\gamma$-rays.

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

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

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

$\therefore$ Total energy given out $=0.0039\times 931MeV$
$=3.6309MeV=3.6309\times {10}^{6}eV$
$=3.6309\times {10}^6\times 1.602\times {10}^{-19}$J
$=5.816\times {10}^{-13}$J
Also, $E=hv$
$5.816\times {10}^{-13}=6.625\times {10}^{-34}\times v$
$\therefore v=8.77\times {10}^{20}Hz$
and $v=c/\lambda$
$\therefore 8.77\times {10}^{20}=(3.0\times {10}^8)/\lambda$
$\therefore \lambda =34\times {10}^{-14}m$