The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength.

Explanation

Here `lamda=589nm=589xx10^(-9)`m
The de Broglie wavelength of a particle having energy R is given by
`lamda=(h)/(sqrt(2mE)`
or `E=(h^2)/(2mlamda^2)`
A. For electron: Here, `lamda=589nm=589xx10^(-9)`m and `m=9.1xx10^(-27)`kg
`(h^2)/(2mlamda^2)=((6.62xx10^(-34)))/(2xx9.1xx10^(-31)xx(589xx10^(-9))^2)`
`=6.94xx10^(-25)`J
B. For neutron: Here `lamda=589nm=589xx10^(-9)`m and `m=1.66xx10^(-27)J`
`E=(h^2)/(2mlamda^2)=((6.62xx10^(-34))^2)/(2xx1.66xx1-^(-27)xx(589xx10^(-9))^2)`
`=3.81xx10^(-28)J`