Question

If a hydrogen atom is excited from an n=1 state to an n=4 state, how much energy does this correspond to? Is this an absorption or an emission? What is the wavelength of the photon involved in this process? To what region of the electromagnetic spectrum does this correspond?

Answer 1

To determine the wavelength, we use

1/ lamda = R (1/n_f² - 1/n_i²)

where R = 1.097 x 10⁷ m-1, n_i = 1 , and n_f = 4 , we find .

1/ = 1.097 x 10⁷ m (1/16 - 1)

       = - 1.097 x 10⁷ m (15/16)

       = 1.03 x 10^7 m

      = 0.97 x 10^-7 m

          = 97nm

Energy E = hc /lamda

      h= 6.625X10^-34 J.sec

c= 3x10^8 m.sec-1

lamda = 97nm

E = 6.625X10^-34 J.sec x 3x10^8 m.sec-1 / 97x10^-9m

   = 0.205 x 10^-17 J

    = 205 x 10^-20 J

Here the transition to occur energy will be gained by the atom.

X-rays : wavelength 10^-12 m -10^-8 m   range refers to X-rays.