If an electron in `n = 3` orbit of hydrogen atom jumps down to `n = 2` orbit, the amount of energy reeased and the and the wavelength of radiation emitted are

Options

A. `0.85 eV, 6566 Å`
B. `0.89 eV, 1240 Å`
C. `1.89 eV, 6566 Å`
D. `1.5 eV, 6566 Å`

Answer

C

Explanation

`(1)/(lambda) = R[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
`(1)/(lambda) = R ((1)/(2^(2)) - (1)/(3^(2)))`
implies `lambda = (36)/(5 R) = (36 xx 10^(-7))/(5 xx 1.097) = 6.566 xx 10^(-7)`
`lambda = 6566 Å`
`E = (h c)/(lambda) = (6.64 xx 10^(-34) xx 3 xx 10^(8))/(6.566 xx 10^(-7)) = 3.05 xx 10^(-19) J`
`:. E = 1.89 e V`