Question

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 :

• A. 1220 A
• B. 1350
• C. 610
• D. 570

Answer 1

Answer: (A)

1220 A

$E_1=-2.17\times {10}^{-11}/1=-2.17\times {10}^{-11}$ erg
$E_2=-2.17\times {10}^{-11}/2^2=-0.5425\times {10}^{-11}$ 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${}^8$ = 1220.82 A