QJEEM21Phdce2: JEE Main 2021 Physics - Class 12 JEEM21Phdce2 Solution: The de-Broglie wavelength of a particle having kinetic energy E …

Question

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The de-Broglie wavelength of a particle having kinetic energy E is

\lambda

. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

Accepted Answer

\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}

,

mv = \sqrt {2mE}

\lambda \propto {1 \over {\sqrt E }}

{{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}} = {3 \over 4}

,

{\lambda _2} = 0.75{\lambda _1}

{{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}} \right)^2}

{E_2} = {{16} \over 9}{E_1} = {{16} \over 9}E

(E₁ = E)

Extra energy given =

{{16} \over 9}E - E = {7 \over 9}E

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