Electrons in a sample of H-atoms make transition from state n=x to some lower excited state. The emission spectrum from the sample is found to contain only the line belonging to a particular series. If one of the photons has an energy of 0.6375 eV . Find the value of x.
`["Take" 0.6375 eV = 3/4 xx0.85 eV]`

Answer

x=8

Explanation

We have
`DeltaE=3/4xx0.85eV`
as energy=0.6375 the photon will belong to brackett series (as for brackett`0.31leEle0.85`)
`0.85xx(1-1/4)=13.6(1/4^2-1/n^2)`
`0.85(1-1/4)=13.6/16[1/(4/n)^2] :. 4/n=1/2 implies n=8` Hence x=8