Calculate the energy emitted when electrons of `1.0 g` of hydrogen undergo transition giving spectrum lines of the lowest energy in the visible region of its atomic spectrum.
`R_(H) = 1.1 xx 10^(7) m^(-1) , c= 3 xx 10^(8) m s^(-1)` and `h = 6.62 xx 10^(-34) J s`

Explanation

For visible line spectrum , i.e. Boalumer series ` n_1 =2`1 Also for minimum energy transition ` n_2 =3` .
` :. 1/( lambda) = R_H [ 1/n_1^2 - 1/n_2^2]`
` :. 1/(lambda) = RH [ 1/2^2 - 1/3^2 ]`
` -= 1.1 xx 10^7 [ 1/4 - 1/9] = 1.1 xx 10^7 xx 5/( 36)`
` :. lambda = 6. 55 xx 10^(-7) ` meter
Now ` E= (hc)/( lambda) = ( 6.63 xx 10 10^(-34) xx 3 xx 10^8)/( 6. 55 xx 10^(-70)`
` = 3 . 03 xx 10 ^(-19) ` joule
If (N) electrons shown this transition in 1 g-atom of (H) then
Energy released ` = E xx N`
` = 3. 03 xx 10 ^(-19) xx 6.023 xx 10^(23)`
` = 18 . 25 xx 10^4 J = 182 . 5 k J`.