What is the maximum value of the wavelength of light that will ionize the hydrogen atom? Express your answer in units of nanometers.
1/λ =R(1/n² - 1/m²) =
=1.1•10⁵(1/1²-1/∞)
λ = 9.1•10⁻⁶ m
To determine the maximum value of the wavelength of light that will ionize the hydrogen atom, we can use the Rydberg formula. The formula is:
1/λ = R(1/n₁² - 1/n₂²)
Where:
- λ is the wavelength of light
- R is the Rydberg constant, approximately 1.097 x 10^7 m^-1
- n₁ and n₂ are positive integers representing the energy levels of the hydrogen atom.
In this case, since we want to find the value that will ionize the hydrogen atom, we set n₁ = 1 and n₂ = ∞ (since the final energy level for ionization is infinite).
Substituting these values into the formula, we get:
1/λ = R(1/1² - 1/∞²)
1/λ = R - 0
1/λ = R
Rearranging the equation to solve for λ, we have:
λ = 1/R
Substituting the numerical value for R (1.097 x 10^7 m^-1), we can find the maximum value of λ. However, to express it in nanometers, we need to convert from meters to nanometers by multiplying by 10^9:
λ = 1/(1.097 x 10^7 m^-1)
λ = 9.12 x 10^-8 m
λ = 9.12 x 10^-8 m x 10^9 nm/m
λ = 9.12 x 10^1 nm
λ ≈ 91.2 nm
Therefore, the maximum value of the wavelength of light that will ionize the hydrogen atom is approximately 91.2 nanometers.
To find the maximum value of the wavelength of light that will ionize the hydrogen atom, we need to use the formula for the energy of a photon. The formula is given by:
E = hc/λ
Where:
- E is the energy of the photon
- h is Planck's constant (approximately 6.626 x 10^-34 J·s)
- c is the speed of light (approximately 3.0 x 10^8 m/s)
- λ is the wavelength of the photon
For ionization to occur in the hydrogen atom, the energy of the photon needs to be equal to or greater than the ionization energy of hydrogen, which is approximately 13.6 eV (electron volts), or 2.18 x 10^-18 J.
Since we want to find the maximum value of the wavelength, we want to find the minimum value of the energy, which occurs when E = 13.6 eV.
So, we can rewrite the formula for energy as:
13.6 eV = hc/λ
To convert eV to Joules, we use the conversion factor: 1 eV = 1.6 x 10^-19 J. Therefore, 13.6 eV is equal to 2.18 x 10^-18 J.
Now we can rearrange the equation to solve for the maximum value of the wavelength (λ):
λ = hc/(13.6 eV)
Substituting the values for h and c, we have:
λ = (6.626 x 10^-34 J·s)(3.0 x 10^8 m/s)/(2.18 x 10^-18 J)
Simplifying the expression:
λ = (1.9878 x 10^-25 J·m)/(2.18 x 10^-18 J)
λ ≈ 9.10 x 10^-8 m
To express the result in nanometers, we can convert meters to nanometers by multiplying the value by 10^9:
λ ≈ 9.10 x 10^-8 m * 10^9 nm/m
λ ≈ 91.0 nm
Therefore, the maximum value of the wavelength of light that will ionize the hydrogen atom is approximately 91.0 nanometers.