A proton and an <math> <mi>α</mi> </math> -particle both initially at rest are accelerated in a region of electric potential difference V and gravity free space. The proton is projected with gained kinetic energy against a uniform constant electric field and comes to momentary rest after travelling a distance <math> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> . If the <math> <mi>α</mi> </math> -particle is also projected with gained kinetic energy against same electric field, it will come to momentary rest after travelling through:

Gain in K.E. = Loss in P.E. K p = e V for proton K α = 2 e V for α - particleAgain, Loss in K.E. = Work against field K p = e E ⋅ S 0 for proton ⇒ e V = e E S 0 ∴ S 0 = V E K α = 2 e E . S for α - particle ⇒ 2 e V = 2 e E S ∴ S = V E = S 0

A proton and an <math> <mi>α</mi> </math> -particle both initially at rest are accelerated in a region of electric potential difference V and gravity free space. The proton is projected with gained kinetic energy against a uniform constant electric field and comes to momentary rest after travelling a distance <math> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> . If the <math> <mi>α</mi> </math> -particle is also projected with gained kinetic energy against same electric field, it will come to momentary rest after travelling through: &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

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Physics Class 12th Physics Electrostatic Potential and Capacitance

A proton and an $α$ -particle both initially at rest are accelerated in a region of electric potential difference V and gravity free space. The proton is projected with gained kinetic energy against a uniform constant electric field and comes to momentary rest after travelling a distance $S_{0}$ . If the $α$ -particle is also projected with gained kinetic energy against same electric field, it will come to momentary rest after travelling through:

Option 1 -

$S_{0}$

Option 2 -

$\sqrt[]{5} S_{0}$

Option 3 -

$\frac{S_{0}}{2}$

Option 4 -

$\frac{S_{0}}{2 \sqrt[]{2}}$

2 Views | Posted 3 weeks ago

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 1

Detailed Solution:

Gain in K.E. $=$ Loss in P.E.

$_{}$ $K_{p} = e V$ for proton

$K_{α} = 2 e V$ for $α$ - particle

Again, Loss in K.E. $=$ Work against field

$K_{p} = e E \cdot S_{0}$ for proton

$\Rightarrow e V = e E S_{0}$

$∴ S_{0} = \frac{V}{E}$

$K_{α} = 2 e E . S$ for $α -$ particle

$\Rightarrow 2 e V = 2 e E S$

$∴ S = \frac{V}{E} = S_{0}$

Gain in K.E. <math> <mo>=</mo> </math> <math></math> Loss in P.E.<math><msub><mrow><mrow></mrow></mrow><mrow><mrow></mrow></mrow></msub></math>  <math> <msub> <mrow> <mrow> <mi>K</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> <mo>=</mo> <mi>e</mi> <mi>V</mi> </math> for proton<math></math> <math> <msub> <mrow> <mrow> <mi>K</mi> </mrow> </mrow> <mrow> <mrow> <mi>α</mi> </mrow> </mrow> </msub> <mo>=</mo> <mn>2</mn> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">V</mi> </math>  for <math></math> <math> <mi>α</mi> </math> - particleAgain, Loss in K.E. <math> <mo>=</mo> </math>  <math></math> Work against field<math><msub><mrow><mrow></mrow></mrow></msub></math> <math> <msub> <mrow> <mrow> <mi>K</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> <mo>=</mo> <mi>e</mi> <mi>E</mi> <mo>⋅</mo> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math>  for proton <math> <mo>⇒</mo> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mi>e</mi> <mi>V</mi> <mo>=</mo> <mi>e</mi> <mi>E</mi> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> <math> <mo>∴</mo> <mrow> <mi mathvariant="normal"> </mi> </mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> </mfrac> </math> <math></math> <math> <msub> <mrow> <mrow> <mi>K</mi> </mrow> </mrow> <mrow> <mrow> <mi>α</mi> </mrow> </mrow> </msub> <mo>=</mo> <mn>2</mn> <mi>e</mi> <mi>E</mi> <mo>.</mo> <mi>S</mi> </math>  for <math> <mi>α</mi> <mo>-</mo> </math> <math></math> particle <math> <mo>⇒</mo> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mn>2</mn> <mi>e</mi> <mi>V</mi> <mo>=</mo> <mn>2</mn> <mi>e</mi> <mi>E</mi> <mi>S</mi> </math> <math> <mo>∴</mo> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mi>S</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math>

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