Ionization energy is defined as the minimum energy required to remove an electron from the ground state (n0) to infinity (n∞). Determine the wavelength of radiation required to ionize the hydrogen electron from the n = 2 energy level. Calculate the energy (Joules) associated with this photon. (1 cm-1 = 1.986 x 10-23 J)
http://en.wikipedia.org/wiki/Rydberg_formula
I will be happy to critique your work.
You may use the value you find for R in the next problem (or the value from a text) to solve for lambda, then E = hc/lambda will calculate energy in Joules. As Bob Pursley says, post your work if you get stuck and we will help you through it. h is Planck's constant and c is the speed of light in meter/second.
To determine the wavelength of radiation required to ionize the hydrogen electron from the n = 2 energy level, we can use the Rydberg formula. The Rydberg formula relates the wavelength of the emitted or absorbed radiation to the energy levels of the hydrogen atom.
The Rydberg formula is given as:
1/λ = R * (1/n1^2 - 1/n2^2)
Where λ is the wavelength, R is the Rydberg constant, and n1 and n2 are the energy levels.
For this problem, we are interested in finding the wavelength when n1 = 2 (initial energy level) and n2 is infinity (final energy level).
Using the Rydberg formula, we can rewrite it as:
1/λ = R * (1/2^2 - 1/∞^2)
Since the term 1/∞^2 can be considered negligible, we can simplify further:
1/λ ≈ R * (1/2^2)
1/λ ≈ R/4
Solving for λ, we get:
λ ≈ 4/R
Now, we need to substitute the value of the Rydberg constant, R, to calculate the approximate wavelength of the radiation required to ionize the hydrogen electron from the n = 2 energy level.
The value of the Rydberg constant, R, is approximately equal to 1.097373 x 10^7 m^-1.
Substituting this value into the equation, we get:
λ ≈ 4 / (1.097373 x 10^7 m^-1)
λ ≈ 3.641 x 10^-7 m
So, the approximate wavelength of the radiation required to ionize the hydrogen electron from the n = 2 energy level is 3.641 x 10^-7 meters.
To calculate the energy associated with this photon in Joules, we can use the equation:
E = hc/λ
where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength.
The value of Planck's constant, h, is approximately equal to 6.626 x 10^-34 J s, and the value of the speed of light, c, is approximately equal to 3 x 10^8 m/s.
Substituting these values into the equation, we get:
E = (6.626 x 10^-34 J s * 3 x 10^8 m/s) / (3.641 x 10^-7 m)
E ≈ 1.147 x 10^-18 J
Therefore, the energy associated with this photon is approximately 1.147 x 10^-18 Joules.