How much energy (in eV)does a photon of orange light (λ = 609 nm)have? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)3.27 eV B)0.31 eV C)2.04 eV D)0.68 eV E)5.23 eV

Question

Quizplus · Accepted Answer

First, we need to calculate the energy of the photon using the equation:

E = hc/λ

where:
h = Planck's constant = 6.63 × 10^-34 J·s
c = speed of light = 3.00 × 10^8 m/s
λ = wavelength of orange light = 609 nm = 609 × 10^-9 m

Substituting the values, we get:

E = (6.63 × 10^-34 J·s × 3.00 × 10^8 m/s) / (609 × 10^-9 m)
E = 3.42 × 10^-19 J

Now, we need to convert this energy to electron volts (eV) using the conversion factor:

1 eV = 1.60 × 10^-19 J

So, dividing the energy by the conversion factor, we get:

E = (3.42 × 10^-19 J) / (1.60 × 10^-19 J/eV)
E = 2.14 eV

Therefore, the answer is C) 2.04 eV (rounding off to two significant figures).

How Much Energy (In EV)does a Photon of Orange Light