What was the wavelength of the second photon emitted?

Question

A ground-state H atom absorbs a photon of wavelength 97.25 nm, and its electron attains a higher

energy level. The atom then emits two photos: one of wavelength 486 nm to reach an intermediate

level, and a second to reach the ground state. What was the wavelength of the second photon emitted?

Answer 1

We will use the Rydberg formula to find the higher and intermediate energy levels of the electron. After that, we can proceed to find the energy of the 2nd photon and hence the wavelength of that photon.

Rydberg formula: 1/λ = R(1/n₁² - 1/n₂²)

R = 1.097x10⁷m^-1

λ = 97.25 nm = 97.25x10^-9 m

n1 = 1 (ground state)

n₂ = ?

Solving for n₂: 1/97.25x10^-9 m = 1.097x10⁷m^-1 (1/1² - 1/n₂²)

0.937 = 1 - 1/n₂²

1/n₂² = 0.0626

n₂ = 4

What is the intermediate level for λ = 1281 nm?

1/1281x10^-9 = 1.097x10⁷ (1/n² - 1/5⁵)

0.0712 = 1/n² - 0.0265

1/n² = 0.111

n = 2.7 = 3 This is the intermediate level

Next, we can calculate the energy of the photon with the wavelength of 1281 nm

E = hc/λ

E = (6.626x10^-34 Js)(3x10⁸m/s) / 1281x10^-9 m)

E = 1.55x10^-19 J

Now, calculate energy of the photon in the ground state

E = (6.626x10^-34 Js)(3x10⁸m/s) / 97.25x10^-9 m

E = 2.04x10^-18 J

Thus, the energy of the 2nd photon will be the difference between the two energies calculated above:

E of 2nd photon =2.04x10^-18 J - 1.55x10^-19 J = 1.89x10^-18 J

Finally, now that we have the energy, we can calculate the wavelength of this photon:

E = hc/λ

λ = hc / E = (6.626x10^-34 Js)(3x10⁸ m/s) / 1.89x10^-18 J

λ = 1.05x10^-7 m = 105 nm