Question

Green light emitted by excited mercury vapor corresponds to a particular energy transition in the mercury atom. A more energetic transition might emit

A. red light
B. white light
C. blue light

Answer 1

To find the energy released by an electron in a mercury atom for a 435.8 nm photon, we use E = hc/λ and convert the wavelength to meters, resulting in approximately 4.561 x 10^-19 joules.

Step-by-step explanation:

To determine the amount of energy released by an electron in a mercury atom to produce a photon of light with a wavelength of 435.8 nm, we can use the equation E = hc/λ, where E is the energy of the photon, h is Planck’s constant (6.626 x 10-34 J·s), c is the speed of light (3 x 108 m/s), and λ is the wavelength of the light in meters.

First, we need to convert the wavelength from nanometers to meters:
λ = 435.8 nm = 435.8 x 10-9 m. Next, we plug in the values into the equation:

E = (6.626 x 10-34 J·s)(3 x 108 m/s) / (435.8 x 10-9 m) = 4.561 x 10-19 J.

Therefore, the electron must release approximately 4.561 x 10-19 joules of energy to produce a photon with a wavelength of 435.8 nm.