Determine the energy of a photon with a wavelength of 6.74 ✕ 102 nm.

E = hc/wavelength.

Convert nm to m.

if a hydrogen atom absorbs a photon containing 4.8x10^-19 j of energy, what component of the atom experienced an increase in energy?

To determine the energy of a photon with a given wavelength, we can use the following equation:

E = (hc) / λ

Where:
E is the energy of the photon (in joules, J)
h is Planck's constant, approximately 6.626 × 10^-34 J · s
c is the speed of light, which is approximately 3.00 × 10^8 m/s
λ is the wavelength of the photon (in meters, m)

First, we need to convert the given wavelength from nanometers (nm) to meters (m). Since 1 nm = 1 × 10^-9 m, the wavelength becomes:

λ = 6.74 × 10^2 nm = 6.74 × 10^-7 m

Now, we can substitute the values into the equation:

E = (6.626 × 10^-34 J · s × 3.00 × 10^8 m/s) / (6.74 × 10^-7 m)

Simplifying the equation:

E = (1.98778 × 10^-25 J) / (6.74 × 10^-7 m)

Dividing the numbers:

E = 2.95 × 10^-19 J

Therefore, the energy of a photon with a wavelength of 6.74 × 10^2 nm is approximately 2.95 × 10^-19 J.