Question

Calculate the frequency of the photon emitted when the electron jumps from the n = 3 to the n = 2 energy level.

Answer 1

Using the Rydbergy equation, we have

1 / = R [(1 / n_f²) - (1 / n_i²)]

where, R = Rydberg constant = 1.097 x 10⁷ m^-1

n_f = final state = 2

n_i = initial state = 3

then, we get

1 / = (1.097 x 10⁷ m^-1) [1 / (2)² - 1 / (3)²]

1 / = (1.097 x 10⁷ m^-1) (5/36)

1 / = 1.5236 x 10⁶ m^-1

= 1 / (1.5236 x 10⁶ m^-1)

= 6.56 x 10^-7 m

Therefore, the frequency of a photon emitted which will be given as -

we know that, f = c / [(3 x 10⁸ m/s) / (6.56 x 10^-7 m)]

f = 4.57 x 10¹⁴ Hz