Question

According to Bohr's theory, the electronic energy of hydrogen atom in the Bohr's orbit is given by
E${}_n$ = $\frac{21.76\times {10}^{-19}}{n^2}$ J
The longest wavelength in $A^o$ of light will be needed to remove an electron from the third orbit of the He${}^{+}$:

• A. 2050 A
• B. 2450 A
• C. 2250 A
• D. none of these

Answer 1

The correct option is A 2050 A

E${}_n$ = -$\frac{21.76\times {10}^{-19}}{n^2}$ J
For He${}^{\oplus }$, E${}_n$ = -$\frac{21.76\times {10}^{-19}\times z^2}{n^2}$ J
E${}_3$(for He${}^{\oplus }$ = -$\frac{21.76\times {10}^{-19}\times 2^2}{3^2}$ J = -9.68 $\times$ 10${}^{-19}$ J
Energy required to remove an e${}^{-}$ form the third Bohr orbit of He${}^{\oplus }$
$\Delta E$ = 0 - E${}_3$ = 0 -(-9.68 $\times$ 10${}^{-19}$ J)
=9.68 $\times$ 10${}^{-19}$ J (Since IE is +ve)
Since $\Delta$ E = -$\frac{hc}{\lambda }$ J = 9.68 $\times$ 10${}^{-19}$ J
$\lambda$ = -$\frac{hc}{\Delta E}$ = -$\frac{6.266\times {10}^{-34}Js\times 3\times {10}^{8}m}{9.68\times {10}^{-19}J}$
= 2.05 $\times$ 10$6-7$ m = 2050 $\times$ 10${}^{-10}$ m = 2050 A