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

(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.

Step-by-step 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 =
`(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 =
`(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.