Question

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.

Answer 1

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency of the photon is 2.47 x $10^{25}$ Hz.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

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

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x $10^{-9}$)

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ 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 $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.

10.20(1.6 x $10^{-9}$) = (6.6 x $10^{-34}$ * 3 x $10^{8}$)/ λ

λ = $\frac{1.98*10^{-25} }{1.632*10^{-8} }$

  = 1.213 x $10^{-17}$

Wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x $10^{-8}$  = 6.6 x $10^{-34}$ x f

f = $\frac{1.632*10^{-8} }{6.6*10^{-34} }$

 = 2.47 x $10^{25}$ Hz

Frequency of the emitted photon is 2.47 x $10^{25}$ Hz.