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If the kinetic energy of the particle is increased to 16 times its previous vlaue, the percentage change in the de Broglie wavelength of the particle is

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a

25

b

75

c

60

d

50

detailed solution

Correct option is

de Broglie wavelength, λ=h2mKwhere m is the mass and K is the kinetic energy of the particle. When kinetic energy of the particle is increased to 16 times, then its de Broglie wavelength becomes,λ'=h2m(16K)=14λ2mK=λ4 (Using (i))% change in the de Broglie wavelength =λ-λ'λ×100=1-λ'λ×100=1-14×100=75%

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