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Question

The de-Broglie wavelength and kinetic energy of a particle are 2000 ˚A and 1 eV , respectively. If its kinetic energy becomes 1 MeV, its de-Broglie wavelength will become,

A
1 ˚A
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B
2 ˚A
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C
5 ˚A
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D
10 ˚A
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Solution

The correct option is B 2 ˚A
De-Broglie wavelength of a particle is given as,
λ=hp=h2m(K.E)

Since mass of the particle remains constant
λ1(K.E)

λλ=(K.E)(K.E)=11×106=1103

λ=λ103=2000103=2 ˚A

Hence, (B) is the correct answer.

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