The wavelength of light from the spectral emission line of sodium is 589 nm.Find the kinetic energy at which an electron .

Explanation

Wavelength of radiation of sodium,
`lambda`=589 nm
`therefore lambda =589xx10^(-9)m`
Mass of electron `m_(e)=9.1xx10^(-31)kg`
Mass of neutron `m_(n)=1.67xx10^(-27) kg`
`h=6.63xx10^(-34)Js` 1eV `=1.6xx10^(-19)J`
`implies`de-Beoglie.s wavelength of particle.
`lambda=(h)/(p)`
but `p=sqrt(2mK)`
`therefore lambda.=(h)/(sqrt(2mK))`
`therefore lambda.=(h^(2))/(2mK)`
`therefore lambda.(h^(2))/(2MK)`
`therefore K=(h^(2))/(2mlambda^(2))`
(a)Kinetic energy of electron,
`K_(e)=(h^(2))/(2m_(e)lambda^(2))`
`therefore K_(e)=((6.63xx10^(-34))^(2))/(2xx9.1xx10^(-31)xx(589xx10^(-9))^(2)`
`=(43.9569xx10^(-68))/(6313962.2xx10^(-49))`
`=0.00000696185xx10^(-19)`
`=9.96xx10^(-25)J`
`=(6.96xx10^(-25))/(1.6xx10^(-19)) eV`
`=4.35xx10^(-6)eV`
`~~4.35 mueV`
(b)Kineic energy of neutron,
`K_(n)=(h^(2))/(2m_(n)lambda^(2))`
`K_(n)=((6.63xx10^(-34))^(2))/(2xx1.67xx10^(-27)xx(589xx10^(-9))^(2))`
`=(43.9569xx10^(-68))/(1158716xx10^(-45))`
`therefore K_(n)=0.00003793xx10^(-23)`
`therefore K_(n)=3.79xx10^(-28)`J
`therefore K_(n)=(3.79xx10^(-28))/(1.6xx10^(-19))J`
`therefore K_(n)=2.36895xx10^(-9)`eV
`therefore K_(n)~~2.37 neV`