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If the mass of neutron = 1.7 xx 10^(-27)...

If the mass of neutron = `1.7 xx 10^(-27)` kg, then the de-Broglie wavelength of neutron of energy 3 eV is `(h = 6.6 xx 10^(-34) J s)`

A

`1.4 xx 10^(-11) m`

B

`1.6 xx 10^(-10)m`

C

`1.65 xx 10^(-11)m`

D

`1.4 xx 10^(-10)m`

Text Solution

Verified by Experts

The correct Answer is:
C
`lambda = (0.286)/(sqrt(E("in eV"))) Å`
`= (0.286)/(sqrt(3)) Å`
`= 1.65 xx 10^(-11)m`
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