Question

Find the binding energy of an electron in the ground state of a hydrogen like atom in whose spectrum the third of the corresponding Balmer series is equal to 108.5 nm

• A. 54.4 eV
• B. 13.6 eV
• C. 112.4 eV
• D. None of these

Answer 1

The correct option is A 54.4 eV

Binding energy in the ground state

$E_b=\frac{E_0{Z}^2}{n^2}$

$=\frac{E_0{Z}^2}{1^2}$

$=E_0{Z}^2$

Wave number of third Balmer line is

$v=\frac{E_0{Z}^2}{hc}(\frac{1}{2^2}-\frac{1}{5^2})$

$=\frac{E_0{Z}^2}{hc}.\frac{21}{100}$

Wavelength of third Balmer line

$\lambda =\frac{1}{v}$

$=\frac{hc}{E_0{Z}^2}.\frac{100}{21}$

$=\frac{19.86\ast {10}^{-26}}{E_b}.\frac{100}{21}$

$E_b=\frac{19.86\ast {10}^{-26}}{\lambda }.\frac{100}{21}$

$=\frac{19.86\ast {10}^{-26}}{108.5\ast {10}^{-9}}.\frac{100}{21}$

$=8.7$ X ${10}^{-}18J$

$=\frac{8.7\ast {10}^{-18}}{1.6\ast {10}^{-}19}eV$

$=54.4eV$