Question

The boron-$\mathbf{14}$nucleus (mass: 14.02266 u) "beta decays," spontaneously becoming an electron (mass: 0.00055 u) and a carbon- $\mathbf{14}$nucleus (mass: 13.99995 u). What will be the speeds and kinetic energies of the carbon-$\mathbf{14}$nucleus and the electron? (Note: A neutrino is also produced. We consider the case in which its momentum and energy are negligible. Also, because the carbon-$\mathbf{14}$ nucleus is much more massive than the electron it recoils ''slowly'';$\mathbf{\therefore }{\mathit{\gamma }}_{\mathbf{C}}\mathbf{\cong }\mathbf{1}$ .)

Short answer

After beta decay of boron-$\mathbf{14}$, the carbon-$\mathbf{14}$ atom will have $\mathbf{2}\mathbf{.}\mathbf{76}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{}\mathbf{J}$ of energymoving in the opposite direction of an electron with thevelocity of $\mathbf{1}\mathbf{.}\mathbf{62}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{c}$and the electron will have energy $\mathbf{3}\mathbf{.}\mathbf{29}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{}\mathbf{J}$ with a velocity of $\mathbf{0}\mathbf{.}\mathbf{9997}\mathit{c}\mathbf{.}$

Step-by-step solution

Explain Beta decay

Beta-decay is a nuclear reaction where the atomic number of the parent atom increases by one and the atomic mass number is unchanged.

Beta Decay:

${}_4{}^{14}B⟶{{}_{5}C}^{14}+{{}_{1}e}^0+{\overline{v}}_e\Rightarrow$${{}_{0}n}^1⟶{{}_{1}p}^1+{{}_{1}e}^0+{\overline{v}}_{\theta }$

It implies that one of the neutrons in boron decays into a proton and releases a beta particle which is nothing but an electron and antineutrino.

Apply conservation laws

Now applying both conservation laws,

Momentum ($p=\gamma mu)$conserved:

${\gamma }_B{m}_B{u}_{\theta }={\gamma }_C{m}_C{u}_C{\gamma }_{\theta }{m}_g{u}_{\theta }$

As per the case considered in the question, we will assume the momentum and energy of antineutrino (or a neutrino because mass is the same) is negligible.

Velocity for boron-14 will be zero because it is stationary initially and Lorentz factor for carbon-14, as it is much more massive than the electron it recoils slowly

$\therefore {\gamma }_C\cong 1$.

$0=m_C{u}_C+{\gamma }_e{m}_e{u}_e$

$u_C=-\frac{{\gamma }_{\theta }{m}_{\theta }{u}_e}{m_C}$… (1)

Now to determine the recoil velocity of carbon-14 :

Define the velocity of the electron by applying conservation of energy.

$m_B{c}^2-m_C{c}^2+{\gamma }_{\theta }{m}_{\theta }{c}^2$

${\gamma }_e=\frac{m_B-m_C}{m_e}$

By putting the values of masses given in the problem statement in the above equation and equating it with the formula for the Lorentz factor, we get the velocity of the electron as,

$\frac{1}{\sqrt{1-u_v^2}}=\frac{14.02266\mathrm{u}-13.99995\mathrm{u}}{0.00055\mathrm{u}}$

$\sqrt{\frac{1}{1-\frac{u_2^2}{c^2}}}=41.29$

$\frac{u_e^2}{c^2}=1-\frac{1}{(41.29{)}^2}$

$U_e=0.9997c$

Therefore, we can calculate the velocity for carbon-14 using equation (1),

localid="1659089034595" $\begin{array}{rcl}{u}_c& =& -\frac{1}{\sqrt{1-\frac{{\left(0.9997c\right)}^2}{c^2}}}\left(\frac{0.00055}{13.99995}\right)\left(0.9997c\right)\\ & =& -\frac{1}{\sqrt{1-0.9994}}\left(3.927\times {10}^{-5}\times c\right)\\ & =& -\frac{1}{0.0245}\left(3.927\times {10}^{-5}\times c\right)\\ & =& -1.62\times {10}^{-3}c\\ & & \end{array}$

The recoil velocity is quite slow compared to the velocity of the electron.

Determine kinetic energy of carbon-14 and electron

Consider the known data as below.

Mass of electron, $m_e=9.1\times {10}^{-31}kg$

Speed of light,$c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

For the electron, as the velocity is comparable to the speed of light we must solve for kinetic energy in relativistic terms. And for carbon, as the velocity is quite slow, solve classically.

For electron the kinetic energy is,

$\begin{array}{rcl}K{E}_e& =& \left({\gamma }_e-1\right)m_e{c}^2\\ & =& \left(\frac{1}{\sqrt{1-\frac{{\left(0.9997c\right)}^2}{c^2}}}-1\right)\left(9.1\times {10}^{-31}kg\right)\left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)\\ & =& \left(\frac{1}{0.0245}-1\right)\left(81.9\times {10}^{-15}\right)\\ & =& 3.29\times {10}^{-12}J\\ & & \\ & & \end{array}$

Carbon-14:

$\begin{array}{rcl}K{E}_c& =& \frac{1}{2}\left(13.99995\times 1.67\times {10}^{-27}\mathrm{kg}\right){\left(-1.62\times {10}^{-3}\times 3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2\\ & =& \left(11.689\times {10}^{-27}kg\right)\left(23.6196\times {10}^{10}\right)\end{array}$

$=2.76\times {10}^{-15}\mathrm{J}$

Hence, the carbon atom will have $2.76\times {10}^{15}\mathrm{J}$of energy moving in the opposite direction of an electron with the velocity of$1.62\times {10}^3\mathrm{C}$and the electron will have energy $3.29\times {10}^{12}\mathrm{J}$with a velocity of $0.9997c$