Question:
A radioactive nucleus of mass number $A$, initially at rest, emits an $ \alpha$ - particle with a speed $v$. The recoil speed of the daughter nucleus will be
A radioactive nucleus of mass number $A$, initially at rest, emits an $ \alpha$ - particle with a speed $v$. The recoil speed of the daughter nucleus will be
Updated On: May 15, 2024
- $ \frac{2v}{A + 4} $
- $ \frac{4v}{A + 4} $
- $ \frac{4v}{A - 4} $
- $ \frac{2v}{A - 4} $
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The Correct Option is C
Solution and Explanation
$\alpha$-particle is equivalent to helium nucleus.
The emission of an $\alpha$-particle from the atom of an element reduces its atomic number by $2$ , and mass number by $4$ .
Hence, the radioactive emission is as follows :
${ }_{Z} X ^{A} \xrightarrow{\alpha \text {-particle }}{ }_{Z-2} Y ^{A-4}+{ }_{Z} He ^{4}$ ( $\alpha$-particle)
Also form law of conservation of momentum,
$m \times 0 =m_{y} v_{y}+m_{\alpha} v_{\alpha} $
$=(A-4) v_{y}+4 v $
$\Rightarrow v_{y} =-\frac{4 v}{A-4}$
The emission of an $\alpha$-particle from the atom of an element reduces its atomic number by $2$ , and mass number by $4$ .
Hence, the radioactive emission is as follows :
${ }_{Z} X ^{A} \xrightarrow{\alpha \text {-particle }}{ }_{Z-2} Y ^{A-4}+{ }_{Z} He ^{4}$ ( $\alpha$-particle)
Also form law of conservation of momentum,
$m \times 0 =m_{y} v_{y}+m_{\alpha} v_{\alpha} $
$=(A-4) v_{y}+4 v $
$\Rightarrow v_{y} =-\frac{4 v}{A-4}$
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Concepts Used:
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In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
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Following are the terms related to nucleus:
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