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When an \alpha-particle of mass 'm' moving with velocity '\upsilon' bombards on a heavy nucleus of charge 'Ze', its distance of closet approach from the nucleus depends on m as:

  • Option 1)

    \frac{1} {\text{m}}

  • Option 2)

    \frac{1} {{\sqrt \text{m} }}

  • Option 3)

    \frac{1} {{\text{m}^\text{2} }}

  • Option 4)

    m

 

Answers (1)

Whwn \alpha particle is a distance of closest approach its velocity =0. Hence kinetic energy =0. Initial since \alpha particles is at far distance its potential energy =0.

\thereforeInitial energy =\frac{1}{2}mv^2

Final energy =\frac{1}{4\pi \epsilon _0}.\frac{(zl).(2l)}{r_0}

{r_0} is distance of current of closest approach.

From principle of energy conservation initial energy = Final Energy

\frac{1}{2}mv^2 =\frac{1}{4\pi \epsilon _0}.\frac{2ze^2}{r_0}

{r_0}=\frac{4ze^2}{4\pi \epsilon _0mv^2}

\therefore {r_0}=\frac{1}{m}


Option 1)

\frac{1} {\text{m}}

Correct

Option 2)

\frac{1} {{\sqrt \text{m} }}

Incorrect

Option 3)

\frac{1} {{\text{m}^\text{2} }}

Incorrect

Option 4)

m

Incorrect

Posted by

Vakul

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