Question

a bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be released by an electron in a mercury atom to produce a photon of this light?

Answer 1

Answer:

4.56 × 10^-19 Joules

Explanation:

We are given;

• Wavelength of the wave as 435.8 nm

We are required to calculate the amount of energy released by an electron.

• We know that the speed of the wave, c is 2.998 × 10^8 m/s
• But, c = f × λ , where f is the frequency and λ is the wavelength
• Energy of a wave is given by the formula;

E = hf , where h is the plank's constant, 6.626 × 10^-34 J-s

But, f = c/λ

Therefore;

f = (2.998 × 10^8 m/s) ÷ (4.358 × 10^-7 m)

  = 6.879 × 10^14 Hz

Thus;

Energy = 6.626 × 10^-34 J-s ×6.879 × 10^14 Hz

            = 4.558 × 10^-19 Joules

            =  4.56 × 10^-19 Joules

Therefore, the energy that must be released by the electron is 4.56 × 10^-19 Joules