Hydrogen atom is ground state is excited by mean of monochromatic radiation of wavelength `975Å` How many different lines are possible in the resulting spectrum ? Calculate the longest wavelength for hydrogen atom as ionization energy is`13.6 eV`

Answer

`6, lambda_(min)=(16xx9)/(7R)=18800Å`

Explanation

`lambda=975Å`
`E=(hc)/(lambda)`
`=(6.6xx10^(-34)xx3xx10^(8))/(975xx10^(-10)xx1.6xx10^(19))`
`0.0126xx10^(3)eV`
`=12.6 eV`
If `n` is the highest state,
`12.6=13.6((1)/(1^(2))-(1)/(n^(2)))`
`rArr` solving `n=4`
Total no. of transitions `((n-1)xxn)/(2)=6`
longest wave length meets for `n=4` to `n=3`
`(1)/(lambda_(max))R((1)/(3^(3))-(1)/(4^(2)))=R((16-9)/(9xx16))=(7R)/(16xx9)`
`:. lambda_(max)=(16xx9)/(7R)`