Question

A proton (mass $1.67 \times {10^{ - 27}}kg$) striking a neutron (mass nearly equal to the proton) forms a deuteron. What would be the velocity of the deuteron if it is formed by a proton moving left with a velocity of $7.0 \times {10^6}m{s^{ - 1}}$ and a neutron moving right with a velocity $4.0 \times {10^6}m{s^{ - 1}}$.

Answer 1

Hint: An isolated system is a system with no external force acting on it. That is before the collision is equal to after collision. And the rate of change of momentum is directly proportional to the force acting on it along the straight line.

Formula used:
The law of conservation of linear momentum,
Initial momentum of proton$ = - {m_p}{v_p}$
Initial momentum of neutron$ = {m_n}{v_n}$
Then total momentum,
${P_1} = - {m_p}{v_p} + {m_n}{v_n}$ we have, ${m_p} = {m_n}$
$ \Rightarrow {P_1} = - {m_p}{v_p} + {m_p}{v_n}$
$ \Rightarrow - {m_p}({v_p} - {v_n})$
Where ${P_1}$ is the momentum, ${m_p}$is the mass of the proton, ${v_p}$is the velocity of the proton, ${m_n}$ is the mass of the neutron, ${v_n}$is the velocity of the neutron.
The total momentum can be obtained by adding the mass and velocity of the proton and neutron respectively.

Complete step by step answer:
Given, a proton striking a neutron forms a deuteron. That is,
$ \Rightarrow {}_1^1H + {}_0^1n \to {}_1^2H$
Mass of proton, ${m_p} = 1.67 \times {10^{ - 27}}kg$ that is also nearly equal to the mass of neutron therefore we can write,
$ \Rightarrow {m_p} = {m_n}$
$ \Rightarrow 1.67 \times {10^{ - 27}}kg$
Mass of deuteron ${m_d} = 2 \times {m_p}$
A proton is moving left that is moving along negative X-axis the,
Velocity of proton,${v_p} = - 7.0 \times {10^6}m{s^{ - 1}}$
And a neutron is moving along positive X-axis direction (right) then, velocity of neutron, ${v_n} = + 4.0 \times {10^6}m{s^{ - 1}}$
According to the law of conservation of linear momentum, we know that the,
Initial momentum of proton$ = - {m_p}{v_p}$
Initial momentum of neutron$ = {m_n}{v_n}$
Then total momentum,
$ \Rightarrow {P_1} = - {m_p}{v_p} + {m_n}{v_n}$ we have, ${m_p} = {m_n}$
$ \Rightarrow {P_1} = - {m_p}{v_p} + {m_p}{v_n}$
We can take out $ - {m_p}$ since it is a common value
 $ \Rightarrow - {m_p}({v_p} - {v_n})$
Now substitute the values of velocities we get,
${P_1} = - {m_p}(7 - 4) \times {10^6} - 3{m_p} \times {10^6}kgm{s^{ - 1}}$ ……………………. (1)
Now after the striking of the neutron, we get a deuteron.
Let ${m_d}$ be the mass of the deuteron and ${v_d}$ be the velocity of the deuteron.
Then, total momentum of the deuteron$ = {m_d}{v_d}$
$ \Rightarrow {P_2} = 2{m_p}{v_d}$ ……………….(2) ( we have mass of deuteron ${m_d} = 2 \times {m_p}$ )
Now compare (1) and(2) we get,
$ \Rightarrow {P_1} = {P_2}$
$ \Rightarrow - 3{m_p} \times 10 = 2{m_p}{v_d}$
Rearrange the above equation
$ \Rightarrow {v_d} = \dfrac{{ - 3}}{2} \times {10^6}$
$\therefore - 1.5 \times {10^6}m{s^{ - 1}}$

Therefore, the velocity of the deuteron is $ - 1.5 \times {10^6}m{s^{ - 1}}$.

Additional information:
There are three types of decay process-
(i) Alpha decay
(ii) Beta decay
(iii) Gamma decay
The process of emission of an electron or positron from a radioactive nucleus is called $\beta - {\text{decay}}$. When a nucleus emits a $\beta - $ particle (electron or positron), the mass number A of the nucleus does not change but its atomic number Z increases by 1. Therefore, $\beta - {\text{decay}}$can be represented as:
\[ \Rightarrow {\text{X}}_Z^A \to {\text{Y}}_{Z + 1}^A + e_{ - 1}^0 + Q\]
Where Q represents the maximum kinetic energy with which the $\beta $- particles are emitted.$ \Rightarrow Q = \left( {{m_x} - {m_y}} \right){c^2}$
According, the above relation all $\beta $-particles must be emitted with the same kinetic energy. But the experiments showed that only a few $\beta $-particles are emitted with the maximum kinetic energy. A large number of $\beta $- particles are emitted with a small value of kinetic energy. Thus, there is an apparent break or violation of the law of conservation of mass-energy. To overcome this difficulty Pauli proposed a theory called the “neutrino hypothesis”.

Note:
There must exist another particle called “neutrino” to account for the missing energy and momentum. Therefore, in a negative $\beta $ -decay, a neutron transformed into a proton, electron, and an antineutrino. That is, $n \to p + e_{ - 1}^0 + \overline \nu $. Similarly, in a positive $\beta $ -decay, a proton is transferred into a neutron, positron, and a neutrino. That is, $p \to n + e_{ + 1}^0 + \nu $. Neutrinos are neutral particles with very small mass compared to the electron. They have only weak interactions with other particles. Therefore, they are very difficult to detect.