A monochromatic laser is exciting Hydrogen atoms from the n=2 state to the n=5 state.

A) What is the wavelength λ of the laser? Express your answer to three significant digits in nanometers.

(B) Eventually, all of the excited hydrogen atoms will emit photons until they fall back to the ground state. How many different wavelengths can be observed in this process?

(C) What is the longest wavelength λ_max that is observed? Express your answer to three significant digits in nanometers.

(D) What is the shortest wavelength λ_min observed? Express your answer to three significant digits in nanometers.

Answer:- The energy of the laser must be higher than or equal to the energy of the transition of the hydrogen atom, whose levels of energy are defined by the Bohr model, in order to carry out this excitation.

En = -13,606 / n² [eV]

as a result, the transition’s energy is

ΔE = E₅ -E₂

ΔE = 13.606 (1 / n₂² – 1 / n₅²)

ΔE = 13.606 (1/2² – 1/5²)

ΔE = 2,85726 eV

Let’s apply Planck’s equation immediately.

E = h f describes how wavelength and frequencies affect the speed of light.

c = λ f

f = c /λ

E = h c /λ

λ = h c / E

Bring the energy down to the SI system.

E = 1.6 1019 J / 1 eV / 2,85726 eV = 4.5716 1019 J

Let’s compute.

λ = 6,626 10⁻³⁴ 3 10⁸ / 4,5716 10⁻¹⁹

λ = 4.348 10⁺⁷ m (10⁹ nm / 1 m)

From this state with n = 5 to the base state n = 1, photon emission processes at = 435 nm can produce the transitional starting state n = 5.

final condition n = 4

ΔE = 13.606 (1/4² – 1/5²)

ΔE = 0.306 eV

λ = h c / E

λ = 4052 nm

n = 5

the final energy (E) (nm)

level

4 0.306 4052

3 0.9675 1281

2 2,857 435

1 13.06 95

n = 4

3              0.661      1876

2              2,551       486

1              11,905       104

n = 3

2              1.89         656

1             12.09         102.5

n = 2

1               10.20         121.6

1. c) λ = 4052 nm
2. d) λ= 95 nm