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  • The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of.The ground state energy of a

The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of108.5 \mathrm{~nm}.The ground state energy of an electron of this ion will be 
 

Option: 1

3.4 \mathrm{eV}


Option: 2

13.6 \mathrm{eV}


Option: 3

54.4 \mathrm{eV}


Option: 4

122.4 \mathrm{eV}


Answers (1)

best_answer

For third line of Balmer series \mathrm{n_1=2, n_2=5}
\mathrm{\therefore \frac{1}{\lambda}=R Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\: gives\: Z^2=\frac{n_1^2 n_2^2}{\left(n_2^2-n_1^2\right) \lambda R}}

On putting values \mathrm{Z= 2}
\mathrm{From\: E=-\frac{13.6 Z^2}{n^2}=\frac{-13.6(2)^2}{(1)^2}=-54.4 \mathrm{eV}}

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