Question

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula: dll In this equation R, stands for the Rydberg energy and stands for the principal quantum number of the orbital that holds the electron. (You can find the value of the Rydberg energy using the Data button on the ALEKS toolbar.) Calculate the wavelength of the line in the absorption line spectrum of hydrogen caused by the transition of the electron from an orbital with a = 4 to an orbital with 9. Round your answer to 3 significant digits ? Con

Answer 1

The wavelength of absorption is the wavelength corresponding to the energy difference between two states.

The energy difference(ΔE) between two states n1 and n2 is given by:

$\Delta E=R\times \left (\frac{1}{n_1^2}-\frac{1}{n_2^2} \right )$

where R is the Rydberg constant

R=2.178 *10^(-18) J

The energy difference(ΔE) and wavelength λ are related as follows:

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

OR

$\frac{1}{\lambda}=\frac{ \Delta E}{hc}$

where h is the Planck's constant and c is the speed of light.

Therefore we have:

$\frac{1}{\lambda}=\frac{R}{hc}\times \left (\frac{1}{n_1^2}-\frac{1}{n_2^2} \right )$

OR

$\frac{1}{\lambda}=\frac{2.178\times10^{-18} }{6.626\times 10^{-34 }\times2.998\times10^{8}}\times \left (\frac{1}{n_1^2}-\frac{1}{n_2^2} \right )$

OR

$\frac{1}{\lambda}=1.096\times 10^7\times \left (\frac{1}{n_1^2}-\frac{1}{n_2^2} \right )\: \: m^{-1}$

We have:

n1 = 4

n2 = 9

Therefore we have:

$\frac{1}{\lambda}=1.096\times 10^7\times \left (\frac{1}{4^2}-\frac{1}{9^2} \right )\: \: m^{-1}$

OR

$\frac{1}{\lambda}=5.497\times 10^{5}\: \: m^{-1}$

Therefore the wavelength(λ) is:

${\lambda}=\frac{1}{5.497\times 10^{5}\: \: m^{-1}}=182\times 10^{-8}\: \: m$