The energy of the electron in the second and third Bohr's orbitals of the hydrogen atom is `-5.42 xx 10^(-12) erg` and `-2.42 xx 10^(-12) erg` respectively ,Calculate the wavelength of the emitted radiation when the electron drops from the third to the second orbit.

Answer

A, B, C, D

Explanation

`Delta E = - 2.41 xx 10^(-12) - (-5.42 xx 10^(-12))`
` = 3.01 xx 10^(-12) erg`
`Delta E = hc, v = (Delta E)/(h)= (3.01 xx 10^(-12))/(6.62 xx 10^(-27))`
`v=c/lambdaRightarrowlambdac-v=(3xx10^(10))/(3.01xx10^(-12)//(6.62xx10^(-27))) = 6.6 xx 10^(-5)`
`= 6.6 xx 10^(-5) xx 10^(8)`
`= 6.6 xx 10^(3) cm`