A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.03 nm. It then gives off a photon having a wavelength of 1005 nm. What is the final state of the hydrogen atom?
What is the equation to solve this?
find the energy of 93.03 nm
E=plancksconstant*speedlight/lambda
Next, add that to ground state energy (look it up)
then find the energy of the emitted photon as above, then
you have the final energy state.
finalenergystat=groundstate/n^2
solve for n
To determine the final state of the hydrogen atom, you can use the equation for the energy of a photon:
E = hc/λ
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the photon.
First, let's find the energy of the absorbed photon with a wavelength of 93.03 nm:
E1 = (6.626 x 10^-34 Js)(3.00 x 10^8 m/s)/(93.03 x 10^-9 m)
≈ 2.122 x 10^-18 J
Next, let's find the energy of the emitted photon with a wavelength of 1005 nm:
E2 = (6.626 x 10^-34 Js)(3.00 x 10^8 m/s)/(1005 x 10^-9 m)
≈ 1.973 x 10^-18 J
Since energy is conserved, the energy absorbed must equal the energy emitted:
E1 = E2
Therefore, the final state of the hydrogen atom is determined by the energy level associated with an energy of 1.973 x 10^-18 J. To find the corresponding energy level, you can use the equation for the energy of a hydrogen atom:
E_n = -13.6eV/n^2
where E_n is the energy of the nth energy level, -13.6 eV is the ionization energy of hydrogen, and n is the principal quantum number.
Rearranging the equation, we can solve for n:
n^2 = -13.6eV/E_n
Let's substitute the energy value:
n^2 = -13.6eV/(1.973 x 10^-18 J)
To convert eV to joules, 1 eV is equal to 1.602 x 10^-19 J.
n^2 = -13.6eV/(1.973 x 10^-18 J) * (1.602 x 10^-19 J/eV)
≈ 17.55
Taking the square root of both sides, we find:
n ≈ 4.19
The final state of the hydrogen atom is approximately the fourth energy level (n = 4).