Question

The ionization potential of the ground state of a hydrogen atom is $2.17\times {10}^{-11}$ ergs per atom. What would be the wavelength of a photon that is emitted when an electron from the 3rd Bohr orbit returns to the 1st orbit in the hydrogen atom?

• A. $1370\stackrel{o}{A}$
• B. $2737\stackrel{o}{A}$
• C. $1026\stackrel{o}{A}$
• D. $8212\stackrel{o}{A}$

Answer 1

The correct option is C $1026\stackrel{o}{A}$

Given ionization potential is,
$E=-2.17\times {10}^{-11}$ ergs per atom
$=-2.17\times {10}^{-18}Jato{m}^{-1}$
$\therefore \Delta E=E_3-E_1=2.17\times {10}^{-18}(\frac{1}{9}-1)=1.93\times {10}^{-18}Jato{m}^{-1}$
Again,
$\Delta E=\frac{hc}{\lambda }$
$\Rightarrow \lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.93\times {10}^{-18}}=10.26\times {10}^{-8}m=1026\stackrel{o}{A}$