Answer:
The frequency of this radiation 163.83 sec⁻¹ or Hz
Explanation:
The wavelength of a bright line in the visible emission spectrum of mercury vapor is 546.1 nm.
∵1 nm = 10⁻⁹ m
⇒ 546.1 nm = 546,1 x 10⁻⁹ m
We know Frequency(ν) = Speed of light(C) x Wavelength(λ)
⇒ Frequency = 3 x 10⁸ x 546.1 x 10⁻⁹
= 1638.3 x 0.1
= 163.83 sec⁻¹ or Hz
∴ The frequency of this radiation 163.83 sec⁻¹ or Hz
