Question

Determine the wavelengths of all the possible photons that can be emitted from the n = 5 state of a hydrogen atom.

Answer 1

Answer:

Wavelengths of all possible photons are;

λ1 = 9.492 × 10^(-8) m

λ2 = 1.28 × 10^(-6) m

λ3 = 1.28 × 10^(-6) m

λ4 = 4.04 × 10^(-6) m

Step-by-step explanation:

We can calculate the wavelength of all the possible photons emitted by the electron during this transition using Rydeberg's equation.

It's given by;

1/λ = R(1/(n_f)² - 1/(n_i)²)

Where;

λ is wavelength

R is Rydberg's constant = 1.0974 × 10^(7) /m

n_f is the final energy level = 1,2,3,4

n_i is the initial energy level = 5

At n_f = 1,.we have;

1/λ = (1.0974 × 10^(7))(1/(1)² - 1/(5)²)

1/λ = 10535040

λ = 1/10535040

λ = 9.492 × 10^(-8) m

At n_f = 2,.we have;

1/λ = (1.0974 × 10^(7))(1/(2)² - 1/(5)²)

1/λ = (1.0974 × 10^(7))(0.21)

1/λ = 2304540

λ = 1/2304540

λ = 4.34 × 10^(-7) m

At n_f = 3, we have;

1/λ = (1.0974 × 10^(7))(1/(3)² - 1/(5)²)

1/λ = (1.0974 × 10^(7))(0.07111)

1/λ = 780373.3333333334

λ = 1/780373.3333333334

λ = 1.28 × 10^(-6) m

At n_f = 4, we have;

1/λ = (1.0974 × 10^(7))(1/(4)² - 1/(5)²)

1/λ = (1.0974 × 10^(7))(0.0225)

1/λ = 246915

λ = 1/246915

λ = 4.04 × 10^(-6) m