Question

Find the energy required to excite `1.10L` od hydrogen gas at `1.0` nm and `298 K` to the first excited state of atomic hydrogen .The energy required for the dissacitation of `H-H` bond is `436 kJ mol^(-1)`.Also calculate the minimum frequency of a photon to break this bond

Answer 1

Let as first the number of moles of hydrogen atom
`n_(H_2) = (PV)/(KT) = (1 xx 1.10)/(0.0921 xx 298) = 0.045`
Thus the energy required to break `0.045` and of `H_(2)(H-H) "bond")` is `0.045 xx 436 = 19.62 kJ`
Now calculate the energy needed excite the H atom in the first excited sstate i.e. to `n = 2`
`Delta E = 2.18 xx 10^(--18) xx Z^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) J "atom"^(-1)`
`Delta E = 2.18 xx 10^(-18) xx (1)^(2)((1)/(1^(2)) - (1)/(2^(2))) J "atom"^(-1)`
`= 1.635 xx 10^(-18) J "atom"^(-1)`
Number of H atom of `H_(2)` molecules)`xx 2 = (0.05 xx 6.02 xx 10^(23)) xx 2= 6.02 xx 10^(22)`
The energy required to excite the given number of H atom is `6.02 xx 10^(22) xx 1.635 xx 10^(-18) J = 98.43 kJ`
So, the total energy required is
`19.62 + 98.43 = 118.05 kJ`
Now energy required to break a single H-H bond is
`(436 xx 10^(3))/(6.023 xx 10^(22)) = 7.238 xx 10^(-19)` = Energy supplied by the photon
`rArr 7.238 xx 10^(-19) = hv = 6.026 xx 10^(-34)(v)`
`:. v = (7.38 xx 10^(-19))/(6.626 xx 10^(-34))`
`rArr v = 1.09 xx 10^(15) Hz`