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In a Rutherford scattering experiment when a projectile of charge Z1 and mass M1 approaches a target nucleus of charge Z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is [2009]
- a)directly proportional to Z1 Z2
- b)inversely proportional to Z1
- c)directly proportional to mass M1
- d)directly proportional to M1 × M2
Correct answer is option 'A'. Can you explain this answer?
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In a Rutherford scattering experiment when a projectile of charge Z1 a...
The kinetic energy of the projectile is given by
Thus energy of the projectile is directly
proportional to Z1, Z2
proportional to Z1, Z2
Most Upvoted Answer
In a Rutherford scattering experiment when a projectile of charge Z1 a...
The energy of the projectile is not directly given in the problem, so we need to find a relationship between the given variables and the energy. From the conservation of energy and momentum, we can write:
$\frac{1}{2}M_1v_i^2 = \frac{1}{2}M_1v_f^2 + \frac{1}{2}M_2v_t^2 + \frac{1}{4\pi\epsilon_0}\frac{Z_1Z_2e^2}{r_0}$
where $v_i$ is the initial velocity of the projectile, $v_f$ is the final velocity after scattering, $v_t$ is the velocity of the target nucleus, and $e$ is the elementary charge.
We can simplify this expression by noting that the initial and final kinetic energies are related by the scattering angle $\theta$:
$\frac{1}{2}M_1v_i^2 = \frac{1}{2}M_1v_f^2 + \frac{1}{2}M_2v_t^2 + \frac{1}{4\pi\epsilon_0}\frac{Z_1Z_2e^2}{r_0}\left(1-\frac{\sin^2\frac{\theta}{2}}{2}\right)$
If we assume that the target nucleus is much more massive than the projectile ($M_2 \gg M_1$), we can neglect the term $M_1v_f^2$ and simplify further:
$\frac{1}{2}M_1v_i^2 \approx \frac{1}{2}M_2v_t^2 + \frac{1}{4\pi\epsilon_0}\frac{Z_1Z_2e^2}{r_0}\left(1-\frac{\sin^2\frac{\theta}{2}}{2}\right)$
Now, we can see that the energy of the projectile is directly proportional to $Z_1Z_2$, as well as inversely proportional to $r_0$ (since the closer the approach, the larger the potential energy). We can also see that the mass of the projectile ($M_1$) is not a direct factor in the energy expression.
Therefore, the correct answer is (a) directly proportional to $Z_1Z_2$.
$\frac{1}{2}M_1v_i^2 = \frac{1}{2}M_1v_f^2 + \frac{1}{2}M_2v_t^2 + \frac{1}{4\pi\epsilon_0}\frac{Z_1Z_2e^2}{r_0}$
where $v_i$ is the initial velocity of the projectile, $v_f$ is the final velocity after scattering, $v_t$ is the velocity of the target nucleus, and $e$ is the elementary charge.
We can simplify this expression by noting that the initial and final kinetic energies are related by the scattering angle $\theta$:
$\frac{1}{2}M_1v_i^2 = \frac{1}{2}M_1v_f^2 + \frac{1}{2}M_2v_t^2 + \frac{1}{4\pi\epsilon_0}\frac{Z_1Z_2e^2}{r_0}\left(1-\frac{\sin^2\frac{\theta}{2}}{2}\right)$
If we assume that the target nucleus is much more massive than the projectile ($M_2 \gg M_1$), we can neglect the term $M_1v_f^2$ and simplify further:
$\frac{1}{2}M_1v_i^2 \approx \frac{1}{2}M_2v_t^2 + \frac{1}{4\pi\epsilon_0}\frac{Z_1Z_2e^2}{r_0}\left(1-\frac{\sin^2\frac{\theta}{2}}{2}\right)$
Now, we can see that the energy of the projectile is directly proportional to $Z_1Z_2$, as well as inversely proportional to $r_0$ (since the closer the approach, the larger the potential energy). We can also see that the mass of the projectile ($M_1$) is not a direct factor in the energy expression.
Therefore, the correct answer is (a) directly proportional to $Z_1Z_2$.
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In a Rutherford scattering experiment when a projectile of charge Z1 and mass M1 approaches a target nucleus of charge Z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is [2009]a)directly proportional to Z1 Z2b)inversely proportional to Z1c)directly proportional to mass M1d)directly proportional to M1 × M2Correct answer is option 'A'. Can you explain this answer?
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In a Rutherford scattering experiment when a projectile of charge Z1 and mass M1 approaches a target nucleus of charge Z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is [2009]a)directly proportional to Z1 Z2b)inversely proportional to Z1c)directly proportional to mass M1d)directly proportional to M1 × M2Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about In a Rutherford scattering experiment when a projectile of charge Z1 and mass M1 approaches a target nucleus of charge Z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is [2009]a)directly proportional to Z1 Z2b)inversely proportional to Z1c)directly proportional to mass M1d)directly proportional to M1 × M2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a Rutherford scattering experiment when a projectile of charge Z1 and mass M1 approaches a target nucleus of charge Z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is [2009]a)directly proportional to Z1 Z2b)inversely proportional to Z1c)directly proportional to mass M1d)directly proportional to M1 × M2Correct answer is option 'A'. Can you explain this answer?.
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