Question

The polonium isotope ${}_{84}^{210}Po$ is unstable, it emits an $\alpha$ - particle with energy $5.30\text{MeV.}$ The atomic mass of ${}_{84}^{210}Po$ is $209.9829u$ and atomic mass of ${}_2^{4}He$ is $4.0026u.$ Then,

• A. Daughter nucleus is ${}_{82}^{206}Pb$
• B. Disintegration energy of $\alpha$ - decay is $5.30\text{MeV}$
• C. Disintegration energy of $\alpha$ - decay is $5.40\text{MeV}$
• D. Mass of daughter nucleus is $205.9745u$

Answer 1

The correct option is A Daughter nucleus is ${}_{82}^{206}Pb$

By equation balancing we can write the chemical reaction as:
${}_{84}^{210}Po{\to }_{82}^{206}X{+}_2^4+\Delta Q$
Clearly atom $X$ is ${}_{82}^{206}Pb$.

Now for disintegration energy:
By momentum conservation we can write that,
$0=M_l{V}_l-M_{\alpha }{V}_{\alpha }$
$\Rightarrow \frac{M_l}{M_{\alpha }}=\frac{V_{\alpha }}{V_l}$

Total energy of system after disintegration can be written as:
$K.E{.}_{total}=K.E{.}_{Pb}+K.E{.}_{\alpha }$
$=\frac{1}{2}{M}_l{V}_l^2+\frac{1}{2}{M}_{\alpha }{V}_{\alpha }^2$
$=\frac{1}{2}\left(M_{\alpha }{V}_{\alpha }\right)V_l+\frac{1}{2}{M}_{\alpha }{V}_{\alpha }^2$
$=K.E{.}_{\alpha }(1+\frac{V_l}{V_{\alpha }})$
$=K.E{.}_{\alpha }(1+\frac{M_{\alpha }}{M_l})$
$=K.E{.}_{\alpha }\left(\frac{M_l+M_{\alpha }}{M_l}\right)$
$=K.E{.}_{\alpha }\left(\frac{M_{Po}}{M_{Po}-M_{\alpha }}\right)$
Putting respective values we get,
$\therefore$disintegration energy = $5.4eV$

$\Delta E=\Delta m{c}^2$
$5.4MeV=[M_{Po}-M_{Pb}-M_{\alpha }]u\times \frac{931.5MeV}{u}$

Putting respective values, we can easily get value of mass of daughter nucleus as $205.9745u.$