The third line of the Balmer series of an ion | vedclass

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(C) For the third line of the Balmer series,the transition is from $n_2 = 5$ to $n_1 = 2$.Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left[

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$54.4$

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(C) For the third line of the Balmer series,the transition is from $n_2 = 5$ to $n_1 = 2$.Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.Given $\lambda = 108.5 \ nm = 108.5 	imes 10^{-9} \ m$ and $R \approx 1.097 	imes 10^7 \ m^{-1}$.Substituting the values: $\frac{1}{108.5 	imes 10^{-9}} = (1.097 	imes 10^7) Z^2 \left[ \frac{1}{2^2} - \frac{1}{5^2} ight]$.$\frac{1}{108.5 	imes 10^{-9}} = (1.097 	imes 10^7) Z^2 \left[ \frac{1}{4} - \frac{1}{25} ight] = (1.097 	imes 10^7) Z^2 \left[ \frac{21}{100} ight]$.Solving for $Z^2$: $Z^2 \approx \frac{100}{108.5 	imes 10^{-9} 	imes 1.097 	imes 10^7 	imes 21} \approx \frac{100}{25} = 4$.Thus,$Z = 2$.The ground state energy of a hydrogen-like ion is given by $E = -13.6 \frac{Z^2}{n^2} \ eV$.For the ground state,$n = 1$,so $E = -13.6 	imes \frac{2^2}{1^2} = -13.6 	imes 4 = -54.4 \ eV$.The magnitude of the ground state energy is $54.4 \ eV$.

The third line of the Balmer series of an ion equivalent to a hydrogen atom has a wavelength of $108.5 \ nm$. The ground state energy of an electron of this ion will be.....$eV$

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