Consider the following energy levels of a hypothetical atom.

Question

Consider the following energy levels of a hypothetical atom:

E4 −1.21 × 10−19 J 

E3 −5.71 × 10−19 J 

E2 −1.05 × 10−18 J 

E1 −1.55 × 10−18 J 

(a) What is the wavelength of the photon needed to excite an electron from E1 to E4?

(b) What is the energy (in joules) a photon must have in order to excite an electron from E2 to E3?

(c) When an electron drops from the E3 level to the E1 level, the atom is said to undergo emission. Calculate the wavelength of the photon emitted in this process.

Answer 1

ΔE = hc/λ

 

λ = hc/ΔE

 

hc = (6.626 x 10^-34 Js)(3 x 10⁸ m/s) = 1.9878 x 10^-25 Jm

 

λ = (1.9878 x 10^-25)/ΔE, with ΔΕ in J, λ in meters

 

(a)

 

ΔE = E₄ - E₁ = [(-1.21 × 10^-19) - (-1.55 × 10^-18)] J = 1.429 x 10^-18 J

 

λ = (1.9878 x 10^-25)/(1.429 x 10^-18) m = 1.391 x 10^-7 m

 

(b)

 

λ = (1.9878 x 10^-25)/(E₃ - E₂) = ? m

 

Plug in the numbers for E₃ and E₂, and get λ in meters.  You should be ale to finish from here.

 

(c)

 

λ = (1.9878 x 10^-25)/(E₃ - E₁) = ? m

 

It is just a matter of plugging in the energy levels.  You should be ale to finish from here.

 

Answer 2

a) To excite an electron from E₁ to E₄, the energy of the incident photon E_p must be

    Ep    = E₄ − E₁

    hc/λ = −1.21 × 10⁻¹⁹ J −(−1.55 × 10⁻¹⁸ J) = 1.429 10⁻¹⁸ J

 

   So the wavelength of the photon needed to excite an electron from E₁ to E₄ is

        λ  = hc/1.429 × 10⁻¹⁸ J

            = (6.63× 10⁻³⁴ J.s)(3.0 × 10⁸ m/s)/1.429 × 10⁻¹⁸ J

            = 1.39 × 10⁻⁷ m

            = 139 nm

 

b) To excite an electron from E₂ to E₃, the energy of the photon E_p must be

    E_p = E₃ − E₂
         = −5.71 × 10⁻¹⁹ J −(−1.05 × 10⁻¹⁸ J) = 4.34 x 10⁻¹⁹ J

 

c) When an electron drops from the E₃ level to the E₁ level, the energy of the emitted photon must be

    E_p = E₃ − E₁

    Using the same calculations as those in part a, you should have

    λ = 203 nm