If the mass of neutron is `1.7xx10^(-27)` kg , then the de-Broglie wavelength of neutron of energy 3 eV Is………`(h=6.6xx10^(-34)JS)`

Options

A. `1.4xx10^(-11)m`
B. `1.6xx10^(-10)m`
C. `1.65xx10^(-11)m`
D. `1.4xx10^(-10)m`

Answer

C

Explanation

`lambda=(h)/(sqrt(2mE))`
`=(6.6xx10^(-34))/(sqrt(2xx1.7xx10^(-27)xx3xx1.6xx10^(-19)))`
`=(6.6xx10^(-34))/(4.04xx10^(-23))`
`=1.63xx10^(-11)m`
`~~1.65xx10^(-11)m`(nearest value)