Electron in a sample of `H-` atoms make transitions from state `n=x` to some lower excited state. The emission spectrum from the sample is found to contain only the lines belonging to a particular series. If one of the photons had an energy of `0.6375 eV`. Then find the value of `x. [" Take " 0.6375eV=(3)/(4)xx0.85eV]`.

Options

A. 16
B. 24
C. 8
D. 20

Answer

3

Explanation

We have, `" "DeltaE=(3)/(4)xx0.85eV`
as energy `=0.6375` the photon will belong to brakett series ( as for brackett `0.31 le E le0.85)`
`0.85xx(1-(1)/(4))=13.6((1)/(4^(2))-(1)/(n^(2)))`
`0.85(1-(1)/(4))=(13.5)/(16)[1-((4)/(n))^(2)]" ":. (4)/(n)=(1)/(2)" "rArr" "n=8`