91Ó°ÊÓ

We need to find energy is released, in $MeV$, in a${}^{211}Po$decay.

The initial momentum of the polonium isotope is equal to zero since we assume it is at rest initially. So, the products of the decay must have equal and opposite momenta so that the final momentum after the decay (the sum of the momenta of land and particle ) remains zero:

$p_{lead}=p_{\mathrm{Î\pm }}≡p\text{.}$

Now the total energy of a relativistic particle of momentum and mass $m$is $\sqrt{p^2{c}^2+m^2{c}^4}$, while the kinetic energy is this minus the rest energy as:

$T=\sqrt{p^2{c}^2+m^2{c}^4}−m{c}^2$

Rearranging terms:

$\sqrt{p^2{c}^2+m^2{c}^4}=T+m{c}^2$

Squaring this we get as:

$p^2{c}^2+m^2{c}^4=T^2+m^2{c}^4+2m{c}^{2}T$

This yields for the momentum :

$p=\frac{1}{c}\sqrt{T\left(T+2m{c}^2\right)}$

So,

$\begin{array}{rr}p& =\frac{1}{c}\sqrt{T_{lead}\left(T_{lead}+2m_{lead}{c}^2\right)}\\ p& =\frac{1}{c}\sqrt{T_{\mathrm{Î\pm }}\left(T_{\mathrm{Î\pm }}+2m_{\mathrm{Î\pm }}{c}^2\right)}\end{array}$

Equating the right-hand sides of these two equations and squaring the we finding

$T_{lead}\left(T_{lead}+2m_{lead}{c}^2\right)=T_{\mathrm{Î\pm }}\left(T_{\mathrm{Î\pm }}+2m_{\mathrm{Î\pm }}{c}^2\right)$

The last equation can be recast into the quadratic equation for $T_{\mathrm{Î\pm }}$:

$T_{\mathrm{Î\pm }}^2+2m_{\mathrm{Î\pm }}{c}^2{T}_{\mathrm{Î\pm }}−T_{lead}\left(T_{lead}+2m_{lead}{c}^2\right)=0.$

Solving for as:

$T_{\mathrm{Î\pm }}=\sqrt{T_{lead}\left(T_{lead}+2m_{lead}{c}^2\right)+m_{\mathrm{Î\pm }}^2{c}^4}−m_{\mathrm{Î\pm }}{c}^2\text{.}$

In the problem we have that $T_{lead}=0.14\text{MeV}$Also, the mass of the lead nucleus is $m_{lead}=207u$while the mass of the particle is $4\mathrm{μ}$ , where $u=931.49\text{MeV}/{\text{c}}^2$ of mass. Using this data we easily find the kinetic energy of the alpha particle :

$T_{\mathrm{Î\pm }}=7.24\text{MeV}$

So the total energy released is

$T_{\mathrm{Î\pm }}+T_{lead}=7.38\text{MeV}$

Note: This problem is on the borderline of being nonrelativistic so close enough values might be obtained if we use nonrelativistic relation between the energy and the momentum:

$T=\frac{p^2}{2m}⇔p=\sqrt{2mT}$

Then we would have equating the momenta

$\sqrt{2m_{\mathrm{Î\pm }}{T}_{\mathrm{Î\pm }}}=\sqrt{2m_{lead}{T}_{lead}}$

$T_{\mathrm{Î\pm }}=\frac{m_{lead}}{m_{\mathrm{Î\pm }}}{T}_{lead}=\frac{207}{4}{T}_{lead}=7.25\text{MeV}$

Which would yield for the released energy$7.39MeV$.