Question

an electron undergoes a transition from an initial (n_i) to a final (n_f) energy state. The energies of the n_i and the n_f energy states are -2.420*10^-19J and -8.720*10^-20J, respectively. Calculate the wavelength of the light in nanometers corresponding to the energy change of this transition.

Answer 1

Ei = -2.420 x10^-19J and Ef = -8.720 x10^-20J

Now energy difference between them,

E = [nf - ni]

E = [(-8.720 x10^-20) - (-2.420 x10^-19)]

E = 10^-19[(-8.720 x 10^-1 + 2.420)]

E = 10^-19[(-0.8720 + 2.420)]

E = 10^-19[(1.548)]

E = 1.548 x 10^-19 J

relation between energy, speed of light and wavelength can be given as:

E = hc/wavelength

where, h = Plank`s constant = 6.626 x 10^-34Js and c = speed of light = 3 x 10⁸ m/s

Hence, wavelength = hc/E

wavelength = 6.626 x 10 x 10^-34 x 3 x 10⁸ / 1.548 x 10^-19

wavelength = 19.878 x 10^-7 / 1.548

wavelength = 1.284 x 10^-6 meter

as 1 meter = 10⁹ nm,

so value of wavelength in nm = 1.284 x 10^-6 x 10⁹ nm

= 1.284 x 10³ nm