Question

Please show work!!

On an atomic level, when the electron relaxes from a higher energy level (n=4) to a lower energy level, a photon with the wavelength of 485nm is released. Determine the lower energy level, to which the electron relaxes.

Answer 1

Solution:

Given: Energy level occupied by electron initially,n_i = 4 Wavelength of photon released,  = 485nm = 485X10^-19m

Step 1: To calculate energy released by photon,E : We know that, E = h = hc/   Where, h = Planck's constant = 6.636X10^-34 Js c = Speed of light = 3X10⁸ m/s

  E = (6.636X10^-34 Js )(3X10⁸ m/s)/(485X10^-9m)    E = 4.098X10^-19J Now, since energy is being released during the transition from higher level to the lower level, therefore, E​ must be negative. Hence,   E = - 4.098X10^-19J    Step 2: To calculate the lower energy level occupied by electron after relaxation,n_f   E = (2.179X10^-18J)[(1/ n_i²) -(1/n_f²)]    (1/n_f²) =     (1/ n_i²) - {E / (2.179X10^-18J) } (1/n_f²) =    (1/4²) - {  - 4.098X10^-19J   / 2.179X10^-18J)} (1/n_f²) = (1/16) + 0.18807 = 0.25052    n_f²  = 3.99   n_f = 1.99 2

   The lower energy level to which the electron relaxes is n = 2.