Question

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

• A. 3.4 eV
• B. 13.6 eV
• C. 54.4 eV
• D. 122.4 eV

Answer 1

The correct option is C 54.4 eV

Using $\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\Rightarrow \frac{1}{108.5\times {10}^{-9}}=1.1\times {10}^7\times Z^2(\frac{1}{2^2}-\frac{1}{5^2})$
$Z^2=\frac{100}{108.5\times {10}^{-9}\times 1.1\times {10}^7\times 21}=4\Rightarrow Z=2$
Now Energy in ground state $E=-13.6Z^{2}eV=-13.6\times 2^{2}eV=-54.4eV$