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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?

Asked in JEE Main 26th Aug 2nd Shift 2021 · Same or changed energy

Answer: (3) 7/9E

Step-by-step solution

λ∝1/(√E), so (λ')/λ=√E/(E')=0.75.

E'=E/(0.75²)=(16)/9E.

Extra energy =E'-E=(16)/9E-E=7/9E.

Why the other options are wrong

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