Question

What is the energy of a 461 nm photon emitted from a light source?

a) 4.67e-19
b) 2.8e-19
c) 4.31e-19
d) 4.25e-19
e) 3.31e-19
f) 2.92e-19
g) 2.76e-19
h) 4.4e-19
i) 3.74e-19
j) 3.59e-19

Answer 1

The energy of a 461 nm photon emitted from a light source is calculated using Planck's constant and the photon's frequency, which is derived from its wavelength. The answer is found to be 4.31 x 10^-19 J.

Step-by-step explanation:

The energy of a photon is given by the equation E = hf, where h is Planck's constant and f is the frequency of the photon. Frequency can be calculated using the speed of light (c) and wavelength (λ) with the formula f = c / λ. Given the wavelength of the photon is 461 nm (4.61 x 10-7 m), we can use these formulas to find the photon's energy.

First, let's find the frequency of the 461 nm photon:

f = c / λ = (3.00 x 108 m/s) / (4.61 x 10-7 m) = 6.51 x 1014 Hz

Now, we can find the energy:

E = hf = (6.626 x 10-34 J·s) x (6.51 x 1014 Hz) = 4.31 x 10-19 J

Therefore, the energy of a 461 nm photon emitted from a light source is 4.31 x 10-19 J, which corresponds to option (c).