Question

A hydrogen atom initially in its ground state absorbs a photon with a wavelength of 97.25 nm. To which energy level is it excited

Answer 1

This question involves the concept of excitation energy.

The energy level to which the atom is excited is "4".

What is excitation energy?

The excitation energy of an atom is defined as the energy required by the atom to get excited from one energy level to some upper energy level. Mathematically, it can be given by the following formula:

$E=E_o(\frac{1}{n_i^2}-\frac{1}{n_f^2})$

where,

E = excitation energy = $\frac{hc}{\lambda} = \frac{(6.63\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9.725\ x\ 10^{-8}\ m}=2.04\ x\ 10^{-18}\ J$

E (in eV) = $(2.04\ x\ 10^{-18}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})$ = 12.78 eV

h = plank's constant = 6.63 x 10⁻³⁴ J>s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 97.25 nm = 9.725 x 10⁻⁸ m

E₀ = energy of hydrogen atom in ground state = 13.6 eV

$n_i$ = initial orbit = 1

$n_f$ = final orbit = ?

Therefore,

$12.78\ eV=13.6\ eV(\frac{1}{1^2}-\frac{1}{n_f^2})\\\\\frac{12.78\ eV}{13.6\ eV}=1-\frac{1}{n_f^2}\\\\\frac{1}{n_f^2}=1-0.94\\\\n_f^2=\frac{1}{0.06}\\\\n_f=\sqrt{16.67}\\n_f = 4$

Learn more about excitation energy here:

brainly.com/question/2289096

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