Question

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

Answer 1

E${}_1$ = -2.17 $\times$ 10$-11$/1${}^2$ erg = -2.17 $\times$ 10${}^{-11}$ erg

E${}_2$ = -2.17 $\times$ 10$-11$/2${}^2$ erg = -0.5425 $\times$ 10${}^{-11}$ erg

$\Delta$E = E${}_2$ - E${}_1$

= -0.5425 $\times$ 10${}^{-11}$ erg - (-2.17 $\times$ 10${}^{-11}$ erg)

= -1.6275 $\times$ 10${}^{-11}$ erg
$\Delta$ = hv = $\frac{hc}{\lambda }$
$\lambda$ = $\frac{hc}{\Delta E}$ = $\frac{6.62\times {10}^{-27}ergs\times 3\times {10}^{10}cm{s}^{-1}}{1.6275\times {10}^{-11}erg}$

= $\frac{6.62\times {10}^{27}ergs\times 3\times {10}^{10}cm}{1.6275\times {10}^{-11}}$

= 12.2028 $\times$ 10${}^{-6}$ $\times$ 10${}^8$ =$1220.82A^0$