Question

(a) Write the complete $\alpha$ decay equation for ${}^{249}Cf$.

(b) Find the energy released in the decay.

Short answer

(a) The complete $\alpha$ decay equation for ${}^{245}Cf$ is ${}_{98}^{249}C{f}_{151}{\to }_{96}^{245}C{m}_{149}{+}_2^{4}H{e}_2$.

(b) The value for energy released in the decay is $E=6.296\mathrm{M}\mathrm{e}\mathrm{V}$.

Step-by-step solution

Concept Introduction

The process of nuclear decay in which the parent nucleus emits an alpha particle is known as alpha decay. Two protons and two neutrons make up the alpha particle, which is structurally comparable to the nucleus of a helium atom and is symbolised by the Greek letter$\alpha$.

Finding $\alpha$decay equation

(a)

During$\alpha$decay the atomic number decreases by$2$, mass number decreases by$4$, and number of neutrons decreases by$2$. Hence, the decay equation is –

${}_{98}^{249}C{f}_{151}{\to }_{96}^{245}C{m}_{149}{+}_2^{4}H{e}_2$

Therefore, the decay reaction is ${}_{98}^{249}C{f}_{151}{\to }_{96}^{245}C{m}_{149}{+}_2^{4}H{e}_2$.

Energy Released

(b)

Mass of $\mathrm{C}\mathrm{f}$ is: $m_{Cf}=249.0748535\mathrm{u}$.

Mass of $\mathrm{C}\mathrm{m}$ is: $m_{Cm}=245.0654912\mathrm{u}$.

And mass of $\mathrm{H}\mathrm{e}$ atom is: $m_{He}=4.00260325415\mathrm{u}$.

Hence, the mass difference is –

$\begin{array}{c}\delta m=(249.0748535\mathrm{u})-\{(245.0654912\mathrm{u})+(4.00260325415\mathrm{u})\}\\ =0.006759046\mathrm{u}\end{array}$

Hence, the energy released in the decay reaction is –

$\begin{array}{rl}E& =\mathrm{\Delta }m{c}^2\\ E& =(0.006759046\mathrm{u})(931.5\mathrm{M}\mathrm{e}\mathrm{V}/\phantom{\mathrm{M}\mathrm{e}\mathrm{V}\mathrm{u}{\mathrm{c}}^2}\mathrm{u}{\mathrm{c}}^2)\times c^2\\ & =6.296\mathrm{M}\mathrm{e}\mathrm{V}\end{array}$

Therefore, the value for energy is obtained as $E=6.296\mathrm{M}\mathrm{e}\mathrm{V}$.