The energy of the electron in the 2s state is -12.98 eV, the energy in the 3s state -10.84 eV. What is the wavelength of the absorbed photon?
E1 - E2 = h c/(wavelength)
The energy difference (E1 - E2) is 2.14 eV
Change that to Joules and use the equation above.
This is easy stuff. Make an effort.
To find the wavelength of the absorbed photon, we need to apply the equation that relates energy and wavelength in photons.
First, let's find the energy difference between the 2s and 3s states of the electron. We subtract the energy of the 2s state from the energy of the 3s state:
ΔE = Energy of 3s state - Energy of 2s state
ΔE = -10.84 eV - (-12.98 eV)
ΔE = -10.84 eV + 12.98 eV
ΔE = 2.14 eV
Now, we can use the energy-wavelength equation:
E = (hc) / λ
Where:
E is the energy of the photon (in Joules),
h is Planck's constant (6.626 x 10^-34 J·s),
c is the speed of light (3.00 x 10^8 m/s), and
λ is the wavelength of the photon (in meters).
First, we need to convert the energy difference from electron volts (eV) to Joules (J). The conversion factor is:
1 eV = 1.602 x 10^-19 J
So, the energy difference in Joules is:
ΔE (J) = 2.14 eV x (1.602 x 10^-19 J/eV)
ΔE (J) = 3.429 x 10^-19 J
Now, we can rearrange the energy-wavelength equation to solve for the wavelength (λ):
λ = (hc) / ΔE
Substitute the given values:
λ = (6.626 x 10^-34 J·s x 3.00 x 10^8 m/s) / (3.429 x 10^-19 J)
λ = 1.823 x 10^-6 m
Therefore, the wavelength of the absorbed photon is approximately 1.823 x 10^-6 meters.