In H-atom electron jumps from `3^(rd)` to `2^(nd)` energy level, the energy released is -

Options

A. `329.7 kJ`
B. `3.03 xx 10^(-19) J`
C. `182.2 kJ`
D. `145.7 kJ`

Answer

B

Explanation

Energy emitted = `-(1311.8)/(3^2) - (-(1311.8)/(2^2)) kJ mol^(-1)`
or `" " = 1311.8 (1/(2^2) - 1/(3^2)) kJ mol^(-1)`
`= 1311.8 (1/4 - 1/9)`
`= 1311.8 xx 5/36 kJ mol^(-1)`
`:. "Energy emitted per atom " = (1311.8 xx 1^3 xx 5)/(6.02 xx 10^(23) xx 36) J`
`= 3.03 xx 10^(-19) J`