Question

(a)Calculate the energy released if ${}^{238}U$ emits an $\alpha$-particle.

(b) Calculate the energy supplied to ${}^{238}U$ if two protons and two neutrons are to be emitted one by one. The atomic masses of ${}^{238}U$, ${}^{234}Th$ and ${}^{4}He$ are $238.0508u,234.04363u$ and $4.00260u$ respectively.

Answer 1

Given, Atomic mass of ${}^{238}U=238.0508u$

Atomic mass of ${}^{234}Th=234.04363u$

Atomic mass of ${}^{4}He=4.00260u$

To find, a) Energy released if ${}^{238}u$ emits an $\alpha$-particle.

b) Energy to be supplied to ${}^{238}u$ if two protons and two neutrons are emitted one by one.

a) We have, when ${}^{238}u$ emits an $\alpha$-particle, ${}^{238}u\to {}^{234}Th+{}^{4}He$

$\therefore$ Mass defect, $\Delta m=[m.{}^{238}u-(m.{}^{234}Th+m{}^{4}He\left)\right]$

$\Rightarrow \Delta m=[238.0508-(234.0436+4.0026\left)\right]$

$\Rightarrow \Delta m=0.00457u$

So, energy released

$\Rightarrow E=\Delta m{c}^2$

$\Rightarrow E=\left(0.00457u\right)\times 931.5MeV$

$\Rightarrow E=4.25MeV$

b) When two protons and two neutrons are emitted one by one, ${}^{233}u\to {}^{234}Th+2n+2p$

$\therefore$ Mass defect, $\Delta m=m^{238}u-[m{}^{234}Th+2m_n+2m_p]$

$\Rightarrow \Delta m=238.0508u-[234.04363+2\times 1.008665+2(1.007276\left)u\right]$

$\Rightarrow \Delta m=-0.024712u$

$\therefore$ Energy supplied,

$E=\Delta m{c}^2$

$\Rightarrow E=\left(0.024712\right)\times 931.5MeV$

$\Rightarrow E=23.019MeV$

$\Rightarrow E=23.02MeV$

Therefore, a) $4.25MeV$; b) $23.02MeV$