The de-Broglie wavelength of a particle having kinetic energy E is $λ .$ How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

Option 1 -

$\frac{1}{9} E$

Option 2 -

Option 3 -

$\frac{16}{9} E$

Option 4 -

$\frac{7}{9} E$

2 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$λ' = λ \times \frac{75}{100} = \frac{3}{4} λ$

$λ = \frac{h}{p} = \frac{h}{\sqrt[]{2 m E}}$

$\Rightarrow \frac{E'}{E} = {(\frac{λ}{λ'})}^{2} = {(\frac{4}{3})}^{2}$

$\Rightarrow E' = \frac{16}{9} E$

Extra energy required,

$Δ E = E' - E = \frac{7}{9} E$

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The de-Broglie wavelength of a particle having kinetic energy E is $λ .$ How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

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The de-Broglie wavelength of a particle having kinetic energy E is $λ .$ How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?