The graph which shows the variation of $(\frac{1}{λ^{2}})$ and its kinetic energy, $E$ is (where $λ$ is de Broglie wavelength of a free particle):

Option 1 -

(a)

Option 2 -

(b)

Option 3 -

(c)

Option 4 -

(d)

15 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

Correct Option - 4

Detailed Solution:

de-Broglie wavelength $λ = \frac{h}{P} = \frac{h}{m v} = \frac{h}{\sqrt[]{2 m E}}$ where $E = \frac{1}{2} m v^{2}$

Squaring both sides,

$λ^{2} = \frac{h^{2}}{4 m^{2} E}$

$\Rightarrow \frac{1}{λ^{2}} = (constant) E$

Graph passes through origin with constant slope.

de-Broglie wavelength <math> <mrow> <mi>λ</mi> <mo>=</mo> <mfrac> <mrow> <mi>h</mi> </mrow> <mrow> <mi>P</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>h</mi> </mrow> <mrow> <mi>m</mi> <mi>v</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>h</mi> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> <mi>m</mi> <mi>E</mi> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math> where <math> <mrow> <mi>E</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>m</mi> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1816434322"> </o:OLEObject></xml><! [endif]->Squaring both sides, <math> <mrow> <msup> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>h</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>4</mn> <msup> <mrow> <mi>m</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>E</mi> </mrow> </mfrac> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1816434323"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mrow> <mo> (</mo> <mrow> constant </mrow> <mo>)</mo> </mrow> <mi>E</mi> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1816434324"> </o:OLEObject></xml><! [endif]->Graph passes through origin with constant slope.

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The graph which shows the variation of $(\frac{1}{λ^{2}})$ and its kinetic energy, $E$ is (where $λ$ is de Broglie wavelength of a free particle):