
At ${}^{238}U$ nucleus decays by emitting an alpha particle of speed \[v\;m/s\]. The recoil speed of the residual nucleus is (in \[m/s\]).
A. $\dfrac{{ - 4v}}{{234}} \\ $
B. $\dfrac{v}{4} \\ $
C. $\dfrac{{ - 4v}}{{238}} \\ $
D. $\dfrac{{4v}}{{238}}$
Answer
160.8k+ views
Hint: In this question, we are given at ${}^{238}U$ nucleus decays by emitting an alpha particle of speed \[v\;m/s\]. We have to find the recoil speed of the residual of ${}^{238}U$ nucleus. To calculate the speed, we’ll apply the law of conservation of linear momentum.
Formula used:
Momentum – Linear, translational, or simply momentum is the product of an object's mass and velocity in Newtonian mechanics. It is a vector quantity with magnitude and direction.
$P = mv$
where $m$ and $v$ are the mass and speed of the body respectively.
Conservation of total momentum –
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Here, ${m_1},{m_2}$ are the masses of the body, ${u_1},{u_2}$ is the initial velocities of the bodies and ${v_1},{v_2}$ are the final velocities.
Complete step by step solution:
Given that,
Atomic mass of the uranium before decaying, ${m_1} = 238u$
Now, the uranium particle was at rest. Therefore, initial speed is ${v_1} = 0{\text{ }}m/s$
After decaying;
Atomic mass of the alpha particles, ${m_2} = 4u$
Speed of the alpha particle, ${v_2} = v{\text{ }}m/s$
Atomic mass of the uranium nucleus will be, ${m_3} = 238 - 4 = 234u$
Let, the final speed of the uranium be ${v_3} = V$
Applying conservation of linear momentum,
$\text{Momentum before decaying = Momentum after decaying}$
$\Rightarrow {m_1}{v_1} = {m_2}{v_2} + {m_3}{v_3}$
$\Rightarrow 238 \times 0 = 4 \times v + 234 \times V$
$\Rightarrow 4v + 234V = 0$
$\therefore V = \dfrac{{ - 4v}}{{234}}{\text{ }}m/s$
Hence, option A is the correct answer.
Note: The recoil is a result of momentum conservation, as Newton's third law states that the force required to accelerate something will elicit an equal but opposite reaction force, which means that the forward momentum gained by the projectile and exhaust gases will be mathematically balanced. Nuclear reaction, a change in the identity or properties of an atomic nucleus caused by the bombardment of an energetic particle. The particle that is bombarding may be an alpha particle, a gamma-ray photon, a neutron and so on.
Formula used:
Momentum – Linear, translational, or simply momentum is the product of an object's mass and velocity in Newtonian mechanics. It is a vector quantity with magnitude and direction.
$P = mv$
where $m$ and $v$ are the mass and speed of the body respectively.
Conservation of total momentum –
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Here, ${m_1},{m_2}$ are the masses of the body, ${u_1},{u_2}$ is the initial velocities of the bodies and ${v_1},{v_2}$ are the final velocities.
Complete step by step solution:
Given that,
Atomic mass of the uranium before decaying, ${m_1} = 238u$
Now, the uranium particle was at rest. Therefore, initial speed is ${v_1} = 0{\text{ }}m/s$
After decaying;
Atomic mass of the alpha particles, ${m_2} = 4u$
Speed of the alpha particle, ${v_2} = v{\text{ }}m/s$
Atomic mass of the uranium nucleus will be, ${m_3} = 238 - 4 = 234u$
Let, the final speed of the uranium be ${v_3} = V$
Applying conservation of linear momentum,
$\text{Momentum before decaying = Momentum after decaying}$
$\Rightarrow {m_1}{v_1} = {m_2}{v_2} + {m_3}{v_3}$
$\Rightarrow 238 \times 0 = 4 \times v + 234 \times V$
$\Rightarrow 4v + 234V = 0$
$\therefore V = \dfrac{{ - 4v}}{{234}}{\text{ }}m/s$
Hence, option A is the correct answer.
Note: The recoil is a result of momentum conservation, as Newton's third law states that the force required to accelerate something will elicit an equal but opposite reaction force, which means that the forward momentum gained by the projectile and exhaust gases will be mathematically balanced. Nuclear reaction, a change in the identity or properties of an atomic nucleus caused by the bombardment of an energetic particle. The particle that is bombarding may be an alpha particle, a gamma-ray photon, a neutron and so on.
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