Question

H atom in its ground state is excited by means of a radiation of `lambda =970.6` Å. How many different wavelengths are possible in the resulting emission spectrum ?

Answer 1

`lmabda=970.6` Å=97 nm
` DeltaE=(1242)/(lambda(nm))eV=(1242)/(17)=12.75eV`
`DeltaE=13.6 Z^(2)((1)/(I^(2))-(1)/(n^(2)))`
`12.75=13.6(1)^(2)(1-(1)/(n^(2)))`
n=4
Number of wavelengths in resulting spectrum is
`(n(n-1))/(2)=6`
`lambda prop (1)/(DeltaE)`
`Delta E "will be minimum in " to 3`
`(1)/(lambda)=Z^(2)R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=(1)^(2) R((1)/(3^(2))-(1)/(4^(2))=(7)/(144))`
`lambda=(144)/(7R)=(144)/(7xx1.097xx10^(2))=18.75xx10^(-7)m`
`=1875xx10^(-9)m =1875 nm`