Question

Assume an imaginary world where angular momentum is quantized to even multiple. The longest possible wave length emitted by hydrogen in the visible spectrum is

• A. $484nm$
• B. $300nm$
• C. $400nm$
• D. $350nm$

Answer 1

Answer: (A)

$484nm$

The angular momentum is quantized to even multiple of $\overline{h}$, hence
$mvr=2n\hslash$
$mv=\frac{2n\hslash }{r}$
$m{v}^2=\frac{m^2{v}^2}{m}=\frac{(2n\hslash {)}^2}{mr}$

The Coulomb's force of attraction is equal to the centripetal force.
$\frac{Z{e}^2}{4\pi {ϵ}_0{r}^2}=\frac{m{v}^2}{r}$
$\frac{Z{e}^2}{4\pi {ϵ}_0{r}^2}=\frac{\frac{(2n\hslash {)}^2}{mr}}{r}$
$r=\frac{(2n\hslash {)}^{2}4\pi {ϵ}_0}{mZ{e}^2}$

Now, the binding energy of electron is
$BE=\frac{-Z{e}^2}{8\pi {ϵ}_{0}r}$
$BE=\frac{-Z{e}^2}{8\pi {ϵ}_0\left(\frac{(2n\hslash {)}^{2}4\pi {ϵ}_0}{mZ{e}^2}\right)}$
$BE=\frac{-Z^2{e}^{4}m}{32\pi {ϵ}_0^2{n}^2{h}^2}$

$BE=\frac{-3.4}{n^2}eV$

For longest possible wavelength in visible spectrum $n_2=4$ and $n_1=1$
$h\nu =-3.4[1-\frac{1}{4}]$
$h\nu =-3.4\left(\frac{3}{4}\right)$
$\nu =-\frac{3.4\times 0.75}{h}$

Therefore, the longest possible wavelength emitted by hydrogen in the visible spectrum is(in nm)
$\lambda =\frac{c}{\nu }$
$\lambda =\frac{hc}{3.4\times 0.75}$

$\lambda =\frac{1236}{2.55}$

$\lambda =484nm$