Question

A certain radioactive (parent) nucleus transforms to a different (daughter) nucleus by emitting an electron and a neutrino. The parent nucleus was at rest at the origin of an XYcoordinate system. The electron moves away from the origin with linear momentum the neutrino moves away from the origin with linear momentum$\left(1.2\times {10}^{-22}kg.m/s\right)\hat{\mathbf{i}}$ . What are the (a) magnitude and (b) direction of the linear momentum$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\hat{j}$ of the daughter nucleus? (c) If the daughter nucleus has a mass of$\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{26}}\mathbf{kg}$ , what is its kinetic energy?

Short answer

a) The magnitude of the linear momentum of the daughter nucleus,$P=1.4\times {10}^{-22}kg.\frac{m}{s}$ .

b) The direction of the linear momentum of the daughter nucleus, $\theta =28°$ .

c) The kinetic energy of the daughter nucleus,$K.E=1.6\times {10}^{-19}J$ .

Step-by-step solution

Listing the given quantities

The linear momentum of an electron is,$P_e=\left(-1.2\times {10}^{-22}kg.\frac{m}{s}\right)\hat{i}$ .

The linear momentum of the neutrino is,$P_n=\left(-6.4\times {10}^{-23}kg.\frac{m}{s}\right)\hat{j}$ .

Mass of the daughter nucleus is,$m_d=5.8\times {10}^{-26}kg$ .

Understanding the terms momentum, force, kinetic energy

We can find the magnitude of linear momentum of the daughter nucleus using the law of conservation of energy. Using this, we can find the angle made by it with the x-axis, which gives its direction. Then using the formula for K.E. in terms of momentum, we can find the kinetic energy of the daughter nucleus.

(a) Magnitude of the linear momentum of the daughter nucleus

According to the law of conservation of momentum,

The momentum of the system before transformation = Momentum of the system after transformation

Thus, the mathematical formula can be written as,

$\overrightarrow{P_p}=\overrightarrow{P_d}+\overrightarrow{P_e}+\overrightarrow{P_n}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} 0=\overrightarrow{P_d}+\left(-1.2\times {10}^{-22}kg.\frac{m}{s}\right)\hat{i}+\left(-6.4\times {10}^{-23}kg.\frac{m}{s}\right)\hat{j} \\ \overrightarrow{P_d}=\left(1.2\times {10}^{-22}kg.\frac{m}{s}\right)\hat{i}+\left(6.4\times {10}^{-23}kg.\frac{m}{s}\right)\hat{j} \end{gathered}$$

The above expression can be written as,

$\overrightarrow{P_d}=\overrightarrow{P_{d_x}}+\overrightarrow{P_{d_y}}$

The magnitude of the linear momentum of the daughter nucleus is,

$P_d=\sqrt{{\left(\overrightarrow{P_{d_x}}\right)}^2+{\left(\overrightarrow{P_{d_y}}\right)}^2}$

Substitute the values in the above expression, and we get,

role="math" localid="1661321884505" $$\begin{gathered} P_d=\sqrt{{\left(1.2\times {10}^{-22}kg.\frac{m}{s}\right)}^2+{\left(6.4\times {10}^{-23}kg.\frac{m}{s}\right)}^2} \\ =1.36\times {10}^{-22}kg.\frac{m}{s} \\ \approx 1.4\times {10}^{-22}kg.\frac{m}{s} \end{gathered}$$

Therefore, the magnitude of the linear momentum of the daughter nucleus is$1.4\times {10}^{-22}kg.\frac{m}{s}$.

(b) Direction of the linear momentum of the daughter nucleus

The direction of the linear momentum of the daughter nucleus can be given as,

$\theta ={\tan }^{-1}\left|\frac{\left(\overrightarrow{P_{d_y}}\right)}{\left(\overrightarrow{P_{d_y}}\right)}\right|$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\frac{\left(6.4\times {10}^{-23}kg.{\displaystyle \frac{m}{s}}\right)}{\left(1.2\times {10}^{-22}kg.\frac{m}{s}\right)} \\ ={\tan }^{-1}\left(0.533\right) \\ =28° \end{gathered}$$

Therefore, the direction of the linear momentum of the daughter nucleus is$\theta =28°$ .

(c) Kinetic energy of the daughter nucleus

The kinetic energy in the form of momentum is written as,

$K.E.=\frac{P^2}{2m}$

Kinetic energy for the daughter nucleus can be calculated as,

$K.E.=\frac{P_d^2}{2m_d}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} K.E.=\frac{{\left(1.35\times {10}^{-22}\right)}^2}{2\left(5.8\times {10}^{-26}\right)} \\ =1.6\times {10}^{-19}J \end{gathered}$$

Therefore, the kinetic energy of the daughter nucleus is$1.6\times {10}^{-19}J$ .