Question

$\alpha$-particles accelerated by $3\times {10}^5$ volt bombarded a boron target. This results in the nuclear reaction.

${}_2^4\text{He}{+}_5^{10}\text{B}{⟶}_6^{13}\text{C}{+}_1^1\text{H}+\gamma$

$1\times {10}^{5}eV$ energy is used in penetrating the nucleus.If the combined energy of ${}^{13}\text{C}$ and ${}^1\text{H}$ is $5\times {10}^{5}eV$, The energy, frequency and wavelength of $\gamma$-rays is:

$(He=4.0026$ amu, $B=10.0129$ amu, $C=13.0036$ amu)

• A. $5.816\times {10}^{-13}J,8.77\times {10}^{20}Hz,3.4\times {10}^{-13}m$
• B. $5.816\times {10}^{-13}J,7.77\times {10}^{20}Hz,4.4\times {10}^{-13}m$
• C. $6.816\times {10}^{-13}J,9.77\times {10}^{20}Hz,6.4\times {10}^{-13}m$
• D. $7.816\times {10}^{-13}J,10.77\times {10}^{20}Hz,3.4\times {10}^{-13}m$

Answer 1

Answer: (A)

$5.816\times {10}^{-13}J,8.77\times {10}^{20}Hz,3.4\times {10}^{-13}m$

The energy of alpha particle is $E=qV=4\times 3\times {10}^5=1.2\times {10}^{6}eV$

The mass defect is $\Delta m=(4.0026amu+10.0129amu)-(13.0036amu+1.008amu)=3.62\times {10}^{-4}amu$

The binding energy is $3.62\times {10}^{-4}\times 931.48\times {10}^6=3.369\times {10}^{5}eV$

The energy of gamma rays is $1.2\times {10}^{6}eV-(3.369\times {10}^{5}eV+1\times {10}^{5}eV)=3.63\times {10}^{5}eV$

Converting it into joule:

$3.63\times {10}^{5}eV\times 1.602\times {10}^{-19}=5.816\times {10}^{-13}J$

The frequency is $\nu =\frac{E}{h}=\frac{5.816\times {10}^{-13}J}{6.634\times {10}^{-34}}=8.77\times {10}^{20}Hz$

The wavelength is $\lambda =\frac{c}{\nu }=\frac{3\times {10}^{8}m/s}{8.77\times {10}^{20}Hz}=3.4\times {10}^{-13}m$