# Question #2c5a4

Dec 4, 2014

${E}_{2} - {E}_{1} = \Delta E = h \nu = \frac{2 {\pi}^{2} m {Z}^{2} {e}^{4} {K}^{2}}{h} ^ 2 \left[\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right]$

$\Delta E = h \nu$
$\Rightarrow$ 3x${10}^{-} 19$=[(6.626x${10}^{-} 34$)x(3x${10}^{8}$)]/$\lambda$

so $\lambda$=6.626x${10}^{-} 7$m=6626 angstrom

IN DETAIL:
$\nu = \frac{2 {\pi}^{2} m {Z}^{2} {e}^{4} {K}^{2}}{h} ^ 3 \left[\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right]$

now we know that C=$\nu \lambda$
so $\frac{1}{\lambda} = \overline{\nu} = \frac{2 {\pi}^{2} m {Z}^{2} {e}^{4} {K}^{2}}{c {h}^{3}} \left[\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right]$

where $\frac{2 {\pi}^{2} m {e}^{4} {K}^{2}}{c {h}^{3}} = 109678 c {m}^{-} 1 = R$(Rydberg's constant)or 1.09x${10}^{-} 7 {m}^{-} 1$

so our final expression looks like this:
$\overline{\nu} = \frac{1}{\lambda} = R {Z}^{2} \left[\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right]$

ans:

$\frac{1}{\lambda}$=109678[1/4-1/9]

$\lambda$=1/15354.92=6.51x${10}^{-} 5$cm
=6510 angstrom