Saved Bookmarks
| 1. |
The wavelength of light from the spectral emission line of sodium is 589 nm.Find the kinetic energy at which an electron . |
|
Answer» Solution :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` |
|