# One of the lines in the emission spectrum of the hydrogen atom has a wavelength of 102.6 nm. How would you identify the value of n-initial for the transition giving rise to this emission?

Dec 3, 2015

n_("initial")=0.73" "???!!!!!

#### Explanation:

You can use Bohr model where the change on energy is calculated by:

$\Delta E = - 2.178 \times {10}^{- 18} {Z}^{2} \left(\frac{1}{{\left({n}_{\text{final"))^2) - 1/((n_("initial}}\right)}^{2}}\right)$

$\Delta E$ can be calculated from the wavelength by:

$\Delta E = \frac{h c}{\lambda} = \frac{6.626 \times {10}^{- 34} \times 2.998 \times {10}^{8}}{102.6 \times {10}^{- 9}} = 1.94 \times {10}^{- 18} J$

Therefore,
$\implies 1.94 \times {10}^{- 18} = - 2.178 \times {10}^{- 18} \times {\left(1\right)}^{2} \times \left(\frac{1}{{\left(1\right)}^{2}} - \frac{1}{{\left({n}_{\text{initial}}\right)}^{2}}\right)$

$\implies 1.94 \times {10}^{- 18} = - 2.178 \times {10}^{- 18} + \frac{2.178 \times {10}^{- 18}}{{\left({n}_{\text{initial}}\right)}^{2}}$

Solve for ${n}_{\text{initial}} = 0.73$ This is weird. I think there is a problem with the wavelength given.

Anyways, this is how I would solve it.

I hope you find this helpful!