Question

Calculate the wavelength of light emitted from a mol of photons as they transition from in the hydrogen atom.

Answer 1

The question is incomplete, here is the complete question:

How do you calculate the wavelength of the light emitted from a mole of photons as they transition from the n = 4 to the n = 1 principal energy level in the hydrogen atom. Recall that for hydrogen $E_n=-2.18\times 10^{-18}J(\frac{1}{n^2})$

Answer: The wavelength of 1 mole of photons for the given transition is $5.86\times 10^{16}m$

Explanation:

To calculate the change in energy, we use the equation:

$\Delta E=E_1-E_4$

Or,

$E_n=-2.18\times 10^{-18}\left (\frac{1}{n_f^2}-\frac{1}{n_i^2}\right )$

where,

$n_i$ = initial energy level = 4

$n_f$ = final energy level = 1

Putting values in above equation, we get:

$\Delta E=-2.18\times 10^{-18}\left (\frac{1}{1^2}-\frac{1}{4^2}\right )\\\\\Delta E=-2.044\times 10^{-18}J$

To calculate the wavelength of light for 1 mole of photons, we use the equation:

$E=\frac{N_Ahc}{\lambda}$

where,

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light = ?

E = energy emitted =  $2.044\times 10^{-18}J$

Putting values in above equation, we get:

$2.044\times 10^{-18}J=\frac{6.022\times 10^{23}\times 6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{\lambda}\\\\\lambda=\frac{6.022\times 10^{23}\times 6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{2.044\times 10^{-18}J}=5.86\times 10^{16}m$

Hence, the wavelength of 1 mole of photons for the given transition is $5.86\times 10^{16}m$