Part A
Determine the energy of a photon whose absorption changes a He+ ion from the n = 1 to the n = 6 state.
Determine the wavelength of a photon whose absorption changes a He+ ion from the n = 1 to the n = 6 state.
Determine the frequency of a photon whose absorption changes a He+ ion from the n = 1 to the n = 6 state.
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To determine the energy of a photon that changes a He+ ion from the n = 1 to the n = 6 state, we can use the formula:
E = -13.6 eV * (Z^2 / n^2)
Where E is the energy of the photon, -13.6 eV is the energy of an electron at the first energy level (n = 1), Z is the atomic number (in this case, 2 for helium), and n is the principal quantum number for the final state (n = 6).
Substituting the values, we get:
E = -13.6 eV * (2^2 / 6^2)
Simplifying the expression:
E = -13.6 eV * (1/9) = -1.51 eV
Therefore, the energy of the photon is -1.51 eV.
To determine the wavelength of the photon, we can use the formula:
λ = c / ν
Where λ is the wavelength, c is the speed of light (approximately 3.00 × 10^8 m/s), and ν is the frequency of the photon.
Since we have the frequency of the photon, we can use the formula:
ν = E / h
Where ν is the frequency, E is the energy of the photon (in joules), and h is Planck's constant (approximately 6.626 × 10^-34 J⋅s).
Rearranging the formula to solve for E:
E = ν * h
Substituting the values, we get:
E = (-1.51 eV * 1.60 × 10^-19 J/eV) * (6.626 × 10^-34 J⋅s)
Simplifying the expression:
E ≈ -2.47 x 10^-19 J
Now, we can substitute this energy value into the equation for wavelength:
λ = c / ν
λ = (3.00 × 10^8 m/s) / (|E| / h)
Since we are dealing with absolute values, the negative sign can be disregarded.
Substituting the values, we get:
λ = (3.00 × 10^8 m/s) / (2.47 × 10^-19 J / (6.626 × 10^-34 J⋅s))
Simplifying the expression:
λ ≈ 4.83 × 10^-7 m
Therefore, the wavelength of the photon is approximately 4.83 × 10^-7 meters, which corresponds to the visible light range.
Lastly, to determine the frequency of the photon, we can use the energy of the photon calculated earlier:
ν = E / h
Substituting the values, we get:
ν = (-1.51 eV * 1.60 × 10^-19 J/eV) / (6.626 × 10^-34 J⋅s)
Simplifying the expression:
ν ≈ -1.83 × 10^15 Hz
Since frequency cannot be negative, we take the absolute value:
ν ≈ 1.83 × 10^15 Hz
Therefore, the frequency of the photon is approximately 1.83 × 10^15 Hertz.