Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to -13.60 eV.

(a) The energy of the photon is 1.632 x J.

(b) The wavelength of the photon is 1.2 x m.

(c) The frequency of the photon is 2.47 x Hz.

Explanation:

Let;

= -13.60 ev

= -3.40 ev

(a) Energy of the emitted photon can be determined as;

- = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x )

- = 1.632 x Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x Js), c is the speed of light (3 x m/s) and λ is the wavelength.

10.20(1.6 x ) = (6.6 x * 3 x )/ λ

λ =

  = 1.213 x

Wavelength of the photon is 1.2 x m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x  = 6.6 x x f

f =

 = 2.47 x Hz

Frequency of the emitted photon is 2.47 x Hz.

the answer is b, have negligible mass

c. newton's third law of motion