Energy Released for a Photon of Light.

Question

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

Please explain the exponents in detail. Thank you!

Answer 1

The formula you want to use is

E = hv = h*c/λ,

where h=Planck's constant; c = the speed of light (also constant); λ = wavelength of light released

You have two constants and one value which the question provides. All you need to do is plug these values into the equation to solve for E.

Note - you may have convert some units to make sure all the units match up (hint - convert λ)

This is the blueprint -- if you want to work through step by step, please schedule a session with me via instant book!

Answer 2

Not sure what you mean by "explain the exponents in detail", but if you leave a comment, I will try to address the question.

We will use the equation E = hc / λ

E = energy = ?

h = Plancks constant = 6.626x10^-34 J-s

c = speed of light (in vacuum) = 3x10⁸ m/s

λ = wavelength in meters = 435.8 nm x 1 m / 10⁹ nm = 435.8x10^-9 m

Solving for E, we have...

E = (6.626x10^-34 J-s)(3x10⁸ m/s) / 435.8x10^-9 m

E = 4.56x10^-19 J