
$\alpha $ particles are projected towards the nuclei of the different metals, with the same kinetic energy. The distance of closest approach is minimum for:
(A) $Cu\left( {Z = 29} \right) $
(B) $Ag\left( {Z = 47} \right) $
(C) $Au\left( {Z = 79} \right) $
(D) $Pd\left( {Z = 46} \right) $
Answer
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Hint The kinetic energy of the particle is equated with the potential energy because the energy is converted from one form of energy to the other form of the energy and it is given by the law of conservation of the energy. By using this relation of the kinetic energy and the potential energy, then the solution is determined.
Useful formula:
The potential energy between the two charges is given by,
$PE = \dfrac{{k{q_1}{q_2}}}{r} $
Where, $PE $ is the potential energy of the particle, $k $ is the constant value, ${q_1} $ is the charge of the $\alpha $ particle, ${q_2} $ is the charge of the nuclei and $r $ is the distance between the two charges.
Complete step by step answer
Given that,
$\alpha $ particles are projected towards the nuclei of the different metals, with the same kinetic energy.
For the distance of the closest approach, then
$KE = PE\,.....................\left( 1 \right) $
Where, $KE $ is the kinetic energy of the particle and $PE $ is the potential energy of the particle.
By substituting the potential energy formula in the above equation (1), then the equation (1) is written as,
$KE = \dfrac{{k{q_1}{q_2}}}{r} $
By rearranging the terms in the above equation, then the above equation is written as,
$r = \dfrac{{k{q_1}{q_2}}}{{KE}} $
By assuming the $k $, ${q_1} $ and $KE $ all are constant, then the above equation is written as,
$r \propto {q_2} $
From the above equation, as the $r $ value decreases then the ${q_2} $ also decreases, so from the given option, the option (A) is the smaller value compared to the others, so the closest distance is $Cu\left( {Z = 29} \right) $.
Hence, the option (A) is the correct answer.
Note Energy can neither be created nor destroyed. It can be transferred from one form of the energy to the other form of the energy. In this solution also, the kinetic energy is equated with the potential energy, this statement is given by the law of the conservation of the energy.
Useful formula:
The potential energy between the two charges is given by,
$PE = \dfrac{{k{q_1}{q_2}}}{r} $
Where, $PE $ is the potential energy of the particle, $k $ is the constant value, ${q_1} $ is the charge of the $\alpha $ particle, ${q_2} $ is the charge of the nuclei and $r $ is the distance between the two charges.
Complete step by step answer
Given that,
$\alpha $ particles are projected towards the nuclei of the different metals, with the same kinetic energy.
For the distance of the closest approach, then
$KE = PE\,.....................\left( 1 \right) $
Where, $KE $ is the kinetic energy of the particle and $PE $ is the potential energy of the particle.
By substituting the potential energy formula in the above equation (1), then the equation (1) is written as,
$KE = \dfrac{{k{q_1}{q_2}}}{r} $
By rearranging the terms in the above equation, then the above equation is written as,
$r = \dfrac{{k{q_1}{q_2}}}{{KE}} $
By assuming the $k $, ${q_1} $ and $KE $ all are constant, then the above equation is written as,
$r \propto {q_2} $
From the above equation, as the $r $ value decreases then the ${q_2} $ also decreases, so from the given option, the option (A) is the smaller value compared to the others, so the closest distance is $Cu\left( {Z = 29} \right) $.
Hence, the option (A) is the correct answer.
Note Energy can neither be created nor destroyed. It can be transferred from one form of the energy to the other form of the energy. In this solution also, the kinetic energy is equated with the potential energy, this statement is given by the law of the conservation of the energy.
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