The de Broglie wavelength and kinetic en...

The de Broglie wavelength and kinetic energy of a particle is `2000 Å and 1 eV` respectively. If its kinetic energy becomes 1 MeV, then its de Broglie wavelength becomes

A

`1Å`

B

`2Å`

C

`5Å`

D

`10Å`

Verified by Experts

The correct Answer is:

B

`lamda=(h)/(sqrt(2mK))implieslamda_1/lamda_2=sqrt((k_2)/(k_1))`
`implies 2000/(lamda_2)=sqrt((10^6)/(1))implieslamda_2 = (2000)/(1000)=2Å`

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