Question

Calculate the wavelength in Angstroms of the photon that is emitted when an electron in the Bohr orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionization potential of the ground state hydrogen atom is $2.17\times {10}^{-11}$ erg per atom.

• A. $1220\stackrel{\circ }{A}$
• B. $1200\stackrel{\circ }{A}$
• C. $1000\stackrel{\circ }{A}$
• D. $1250\stackrel{\circ }{A}$

Answer 1

The correct option is A $1220\stackrel{\circ }{A}$

Given- $n_1=1,n_2=2$
ionization potential of the ground state hydrogen atom is $2.17\times {10}^{-11}$ erg per atom
ionization potential of the ground state hydrogen atom is $2.17\times {10}^{-18}$ Joule per atom since $1erg={10}^{-7}J$
we know,
$\Delta E=2.17\times {10}^{-18}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{hc}{\lambda }=2.17\times {10}^{-18}[1-\frac{1}{4}]$
$\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{\lambda }=2.17\times {10}^{-18}\times \frac{3}{4}$
$\lambda =\frac{4\times 6.62\times 3\times {10}^{-26}}{6.51\times {10}^{-18}}=\frac{79.44\times {10}^{-26}}{6.51\times {10}^{-18}}=12.20\times {10}^{-8}m=1220\stackrel{\circ }{A}$ [$\because 1m={10}^{+10}\stackrel{\circ }{A}$]
Hence, wavelength of the photon is $1220\stackrel{\circ }{A}$