Conservation of Momentum EX 6

(a) The linear momentum of the radium before the decay is zero. Consequently, the total momentum after the decay is also zero. Therefore, the sum of the momentum of the radon nucleus and the momentum of the $\alpha$-particle is equal to zero

${\overrightarrow{p}}_{Ra}+{\overrightarrow{p}}_{\alpha }$

We deduce from this equation that the momentum of the radon nucleus and the momentum of the $\alpha$-particle have equal magnitudes and opposite directions. Therefore, we write

${\overrightarrow{p}}_{Ra}={\overrightarrow{p}}_{\alpha }$

$222v_{Ra}=4v_{\alpha }$

$v_{Ra}=\frac{4}{222}{v}_{\alpha }$

To compute the recoil speed of the radon nucleus, we need to compute the speed of the $\alpha$-particle from its kinetic energy. To do this we will convert the mass of the $\alpha$-particle from unit $u$ to kg.

$m_{\alpha }=4u\frac{1.66\times 10^{-27}kg}{1u}=6.64\times 10^{-27}kg$

and then compute the speed

$K=\frac{1}{2}{m}_{\alpha }{v_{\alpha }}^2$

$v_{\alpha }=\sqrt{\frac{2K}{m_{\alpha }}}=\sqrt{\frac{2(6.72\times 10^{-13})}{6.64\times 10^{-27}}}=1.42\times 10^{7}m/s$

Then we can compute the speed of the radon nucleus

$v_{Ra}=\frac{4}{222}{v}_{\alpha }=\frac{4}{222}(1.42\times 10^7)=2.56\times 10^{5}m/s$

(b) The kinetic energy of the radon nucleus is

$K_{Ra}=\frac{1}{2}{m}_{Ra}{v_{Ra}}^2=\frac{1}{2}(222\times 1.66\times 10^{-27})(2.56\times 10^5{)}^2=1.21\times 10^{-14}J$