An electronic transition in hydrogen atom result in the formation of `Halpha` line of Hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in `kcal mol^(-1)`) are

Options

A. `-313.6-34.84`
B. `-313.6-78.4`
C. `-78.4-34.84`
D. `-78.4-19.6`

Answer

B

Explanation

Energy of an electron in bth orbit
`E_(n)=(2pi^(2)Z^(2)e^(4))/(n^(2)h^(2))`
On substituting the values of k m e and h we get
`E_(n)=(2.172xx10^(-18)Z^(2))/n^(2)J atom^(-1)`
`=(1311.8Z^(2))/n^(2)kJ mol^(-1)=-(313.52Z^(2))/n^(2)kcal mol^(-1)`
[.: 1 kacal =4.184kJ]
For H - atom
Z=1
For Lyman series
`n_(1)=1 n_(2)=2`
`=-(313.52xx (1)^(2))/(1)^(2)kcalmol^(-1)`
`=-313.52kcal mol^(-1)`
`=-313.6 kcal mol^(-1)`
Energy of electron in `n_(2)` orbit
`=-(313.52xx(1)^(2))/(2)^(2)kcalmol^(-1)`
`-(313.52)/4 kcal mol^(-1)`
`=-78.38 kcal mol^(-1)`