Question

the first line in the lyman series of a spectrum of hydrogen is located at 121.6nm. what is the energy of a photon emitted for the electron transition that gives rise to this line?

Answer 1

Ans. Energy of the photon is given by-

                        E = hv            - equation 1

; where, h = Plank’s constant = 6.626 x 10^-34 Js   ; v = frequency of photon   

Or, E = h x (c / l)      - equation 2

                        Where, c = speed of light    ; l = wavelength

# Given, the first line in the Lyman series is located at 121.6 nm.

That is, the emitted photon has the wavelength of 121.6 nm.

            Or, wavelength, l = 1.216 x 10^-7 m                         ; [1 nm = 10^-9 m]

Using equation 2, the energy of emitted photon is given by-

E = (6.626 x 10^-34 Js) x (299792458 m s^-1 / 1.216 x 10^-7 m)

            Hence, E = 1.6336 x 10^-18 J