Question

What is the maximum wavelength of light capable of removing an electron from a hydrogen atom in the energy states characterized by n=1 and n=3?

Answer 1

To solve this problem we will apply the Rydberg formula is used in atomic physics to describe the wavelengths of the spectral lines of many chemical elements.

This equation is given in its general form as,

$\Delta E = -R_H (\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Here,

$R_H$ = Rydberg constant for Hydroge is approximately $2.178*10^{18} J$

$n_1$= Principal quantum number of an energy level

$n_2$= Principal quantum number of an energy level for the atomic electron transition

PART A ) For n=1 we have that

$\Delta E = -R_H* \frac{1}{n^2}$

$\Delta E = -2.178*10^{-18}*\frac{1}{1^2}$

$\Delta E = -2.178*10^{-18}J$

Now calculating the wavelength using following equation

$\lambda = \frac{hc}{\Delta E}$

Here

h = Planck's constant

c = Speed of light

$\lambda = \frac{(6.626*10^{-24})(2.9979*10^8)}{2.178*10^{-18}}$

$\lambda = 9.120*10^{-8}m = 91.2nm$

PART B) For n = 3 we have that

$\Delta E = -R_H *\frac{1}{n^2}$

$\Delta E = -2.178*10^{-18}*\frac{1}{3^2}$

$\Delta E = -2.42*10^{-19}J$

Now calculating the wavelength using following equation

$\lambda = \frac{hc}{\Delta E}$

$\lambda = \frac{(6.626*10^{-24})(2.9979*10^8)}{2.42*10^{-19}}$

$\lambda = 8.208*10^{-7}m = 820.8nm$