A free particle having one electronic charge with initial kinetic energy <math> <mn>9</mn> <mi> </mi> <mi>e</mi> <mi>V</mi> </math> and de Broglie wavelength <math> <mn>1</mn> <mi> </mi> <mi>m</mi> <mi>m</mi> </math> enters a region of <math> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> potential difference such that new de Broglie wavelength is now <math> <mn>1.5</mn> <mi> </mi> <mi>m</mi> <mi>m</mi> </math> . Then <math> <mi>e</mi> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> is

λ 1 = h 2 m E 1 = λ 2 = h 2 m E 2 = E 2 = 4 9 E 1 = 4 e V E 2 = E 1 - e V 0 = V 0 = 5 V

A free particle having one electronic charge with initial kinetic energy <math> <mn>9</mn> <mi> </mi> <mi>e</mi> <mi>V</mi> </math> and de Broglie wavelength <math> <mn>1</mn> <mi> </mi> <mi>m</mi> <mi>m</mi> </math> enters a region of <math> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> potential difference such that new de Broglie wavelength is now <math> <mn>1.5</mn> <mi> </mi> <mi>m</mi> <mi>m</mi> </math> . Then <math> <mi>e</mi> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> &nbsp;is

Ask & Answer Question

Option 1 -

$5 e V$

Option 2 -

$6 e V$

Option 3 -

$13.5 e V$

Option 4 -

$15 e V$

2 Views | Posted a month ago

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

$λ_{1} = \frac{h}{\sqrt[]{2 m E_{1}}} = λ_{2} = \frac{h}{\sqrt[]{2 m E_{2}}}$

$= E_{2} = \frac{4}{9} E_{1} = 4 e V$

$E_{2} = E_{1} - e V_{0} = V_{0} = 5 V$

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