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
