The wavelength of light from the spectral emission line of sodium is 589nm. Find the kinetic energy at which (a) an electron and (b) a neutron would have the same deBroglie wavelength. Given that mass of neutron`=1.66xx10^(-27)kg`

Explanation

Wavelength of light of a sodium line, `lambda=589 nm =589xx10^(-9) m`
Mass of an electron , `m_e=9.1xx10^(-31) kg`
Mass of a neutron , `m_n=1.66xx10^(-27)` kg
Planck's constant, `h=6.6xx10^(-34) Js`
(a)For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation : `K=1/2 m_e v^2` ....(i)
We have the relation for de Broglie wavelength as :
`lambda=h/(m_e v)`
`therefore v^2=h^2/(lambda^2 m_e^2)`.....(2)
Substituting equation (2) in equation (1), we get the relation :
`K=1/2 (m_e h^2)/(lambda^2 m_e^2)=h^2/(2lambda^2 m_e)`....(3)
`=((6.6xx10^(-34))^2)/(2xx(589xx10^(-9))^2 xx 9.1xx10^(-31))`
`~~6.9xx10^(-25) J`
`=(6.9xx10^(-25))/(1.6xx10^(-19))=4.31xx10^(-6) eV =4.31 mueV`
Hence, the kinetic energy of the electron is `6.9xx10^(-25) J or 4.31 mueV`
(b)Using equation(3), we can write the relation for the kinetic energy of the neutron as : `h^2/(2lambda^2m_n)`
`=((6.6xx10^(-34))^2)/(2xx(589xx10^(-9))^2 xx 1.66xx10^(-27))`
`=3.78xx10^(-28) J`
`=(3.78xx10^(-28))/(1.6xx10^(-19))=2.36xx10^(-9) eV=2.36 n eV`
Hence , the kinetic energy of the neutron is `3.78xx10^(-28) J or 2.36 n eV`