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The de-Broglie wavelength of an electron...

The de-Broglie wavelength of an electron is `0.4 xx 10^(-10)` m when its kinetic energy is 1.0 keV. Its wavelength will be `1.0 xx 10^(-10)`m, When its kinetic energy is

A

0.2 keV

B

0.8 keV

C

0.63 keV

D

0.16 keV

Text Solution

Verified by Experts

The correct Answer is:
D
When de Broglie wavelength of an electron is `0.4 xx 10^(-10)` m then kinetic energy K = 1.0 keV
`because lambda = (h)/(sqrt(2mK)) or 0.4 xx 10^(-10) = (h)/(sqrt(2mK))`
when wavelength is `1 xx 10^(-10)` m , then kinetic energy will be K.
`therefore 1.0 xx 10^(-10) = (h)/(sqrt(2mK))`
On dividing eqn. (i) with eqn, (ii)
`(0.4 xx 10^10)/(1 xx 10^(-10)) = (sqrt(2mK.))/(sqrt(2mK)) or K. = (0.4)^(2) K =0.16 keV.`
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