Question

A photon is emitted as an atom makes a transition from n = 4 to n = 2 level. What is the frequency, wavelength and energy of the emitted photon?

Answer 1

we are given that, Initial orbit of electron = n_i = 4 

Final orbit of electron = n_f = 2. 

Also we know that, Speed of electromagnetic waves = c = 3 × 10⁸ m/s 

Planck’s constant = h = 6.63 × 10⁻³⁴ Js. 

We want to calculate the frequency, wavelength and energy of the emitted photon, when the atom makes a transition from ni = 4 to n_f = 2. 

Frequency of the emitted photon can be calculated by using the equation, 

E = hf, ⇒ f = E/h . 

Let’s at first calculate E by using the following relation,

E = -13.6[1/n²_f - 1/n²_i]eV

= - 13.6[1/2² - 1/4²]eV

= -13.[1/4 - 1/16]eV

= - 13.6 x 3/16eV

= 2.55eV.

Thus frequency of photon will be,

f = E/h = (2.55eV)/(6.63 x 10^-34Js)

= (2.55 x 1.6 x 10^-19J)/(6.63 x 10^-34Js)

= 6.15 x 10¹⁴Hz.

Wavelength of the emitted photon can be calculated by using the following equation.

c = fλ

⇒ λ = c/f = (3 x 10⁸m/s)/(6.15 x 10¹⁴Hz)

= 4.875 x 10^-7m

= 488nm.