Question

A proton is initially at 1A° from an α – particle.

The work done in moving the proton to infinity is​

Answer 1

Given:

A proton is initially at 1A° from an α – particle.

To find:

Work done in moving the proton to $\infty$.

Calculation:

Initial Electrostatic Potential Energy:

$\rm{E1 = \dfrac{k(q1)(q2)}{d } }$

$= > \rm{E1 = \dfrac{k(1.6 \times {10}^{ - 19} )(2 \times 1.6 \times {10}^{ - 19} )}{ {10}^{ - 10} } }$

$= > \rm{E1 = \dfrac{k \times (5.12 \times {10}^{ - 38} )}{ {10}^{ - 10} } }$

$= > \rm{E1 = \dfrac{(9 \times {10}^{9}) \times (5.12 \times {10}^{ - 38} )}{ {10}^{ - 10} } }$

$= > \rm{E1 = 46.08 \times {10}^{19} \times {10}^{ - 38} }$

$= > \rm{E1 = 46.08 \times {10}^{ - 19} \: joule}$

Final Potential Energy:

$\rm{E2 = \dfrac{k(q1)(q2)}{ \infty } }$

$= > \rm{E2 = 0 \: joule }$

So, work done:

$\rm{W = E2 - E1}$

$= > \rm{W = 0 - 46.08 \times {10}^{ - 19} }$

$= > \rm{W = - 46.08 \times {10}^{ - 19} \: joule}$

So, final answer is:

$\boxed{ \bf{W = - 46.08 \times {10}^{ - 19} \: joule}}$