SATHEE: Electrostatics 2 Question 16

Electrostatics 2 Question 16

17. An alpha particle of energy $5 MeV$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of

(a) $1 \AA$

(b) $10^{- 10} cm$

(d) $10^{- 15} cm$

$(1981, 2 M)$

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Answer:

Correct Answer: 17. (c)

Solution:

From conservation of mechanical energy decrease in kinetic energy = increase in potential energy

$\begin{aligned} or \frac{1}{4 π ε_{0}} \frac{(Z e) (2 e)}{r_{min}} = 5 MeV \\ = 5 \times 1.6 \times 10^{- 13} J \\ ∴ r_{min} = \frac{1}{4 π ε_{0}} \frac{2 Z e^{2}}{5 \times 1.6 \times 10^{- 13}} \\ = \frac{(9 \times 10^{9}) (2) (92) {(1.6 \times 10^{- 19})}^{2}}{5 \times 1.6 \times 10^{- 13}} \\ \begin{matrix} (+ Z e) (Z = 92) \end{matrix} \end{aligned}$

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$= 5.3 \times 10^{- 14} m$

$= 5.3 \times 10^{- 12} cm$

i.e. $r_{min}$ is of the order of $10^{- 12} cm$ .

$∴$ Correct option is (c).

Electrostatics 2 Question 16

17. An alpha particle of energy $5 MeV$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of

Answer:

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Electrostatics 2 Question 16

17. An alpha particle of energy 5MeV is scattered through 180∘ by a fixed uranium nucleus. The distance of closest approach is of the order of

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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17. An alpha particle of energy $5 MeV$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of