What wavelength photon would be required to ionize a hydrogen atom in the ground state and give the ejected electron a kinetic energy of 13.1 eV?
I tried ryhdberg's equation and E=hf. Need help.
Thanks
The ground state of a hydrogen atom is 13.6 eV below the ionization threshold. If the free electron is to have 13.1 eV of kinetic energy, the total energy needed is 26.7 eV. Convert that to photon energy E in Joules using 1.6*10^-19 J per eV. Then use the Planck relationship
E = h c/L, where L is the wavelength in meters and h is Planck's constant, to solve for L.
Well, let's see if I can help you put on a happy face. To find the wavelength of the photon, we can start with the energy required to ionize the hydrogen atom, which is 13.1 eV. Now, we need to convert this energy to joules to keep things consistent.
So, 1 electron volt (eV) is equivalent to 1.602 x 10^-19 joules. Multiplying 13.1 eV by this conversion factor, we get 2.097 x 10^-18 joules.
Now, we can use the formula E=hf, where E is the energy, h is Planck's constant (6.626 x 10^-34 Joule·seconds), and f is the frequency. Since we want to find the wavelength, we can use the equation c = fλ, where c is the speed of light (3 x 10^8 m/s) and λ is the wavelength.
By substituting and rearranging the equations, we have c = E/λ, which becomes λ = E/c. Plugging in the values, we get:
λ = (2.097 x 10^-18 J) / (3 x 10^8 m/s)
Calculating this, we find that the wavelength required to ionize the hydrogen atom and give the ejected electron a kinetic energy of 13.1 eV is approximately 6.99 x 10^-10 meters.
Hope that brightens up your day a bit!
To find the wavelength of the photon required to ionize a hydrogen atom in the ground state and give the ejected electron a kinetic energy of 13.1 eV, we can use the energy conservation principle.
1. First, we need to convert the kinetic energy of the ejected electron from electron volts (eV) to joules (J). Recall that 1 eV is equal to 1.602 x 10^-19 J. Therefore, the kinetic energy is:
KE = (13.1 eV) * (1.602 x 10^-19 J/eV)
= 2.0962 x 10^-18 J
2. Next, we can use the energy conservation equation:
E = hf
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of the photon.
3. In the case of a photon, the energy of the photon can also be related to its wavelength (λ) by the equation:
E = hc/λ
where c is the speed of light (3 x 10^8 m/s).
4. By equating the two expressions for E, we get:
hc/λ = hf
5. Solving for frequency (f), we have:
f = c/λ
6. Now we can substitute the obtained expression for frequency (f) back into the energy conservation equation to find the wavelength (λ):
E = hf = hc/λ
λ = hc/E
7. Substituting the values:
λ = (6.626 x 10^-34 J·s) * (3 x 10^8 m/s) / (2.0962 x 10^-18 J)
≈ 9.1 x 10^-7 m
Therefore, the wavelength of the photon required to ionize a hydrogen atom in the ground state and give the ejected electron a kinetic energy of 13.1 eV is approximately 9.1 x 10^-7 meters, or 910 nm.
To solve this problem, we can use the Rydberg formula and the equation for energy of a photon (E = hf) to find the wavelength of the photon required to ionize a hydrogen atom.
Step 1: Find the energy required to ionize the hydrogen atom.
The energy required to ionize a hydrogen atom is equivalent to the ionization energy of hydrogen, which is 13.6 eV.
Step 2: Determine the energy of the photon required to ionize the atom and give the ejected electron a kinetic energy of 13.1 eV.
The total energy of the photon can be found using the equation E = hf, where E is the energy, h is Planck's constant (6.626 × 10^-34 J s), and f is the frequency of the photon. Since energy is given in electron volts, we need to convert it to joules before using the equation. The conversion factor is 1 eV = 1.602 × 10^-19 J.
So, the energy of the photon is:
E = 13.1 eV × (1.602 × 10^-19 J/eV) = 2.095 × 10^-18 J
Step 3: Determine the frequency of the photon.
To find the frequency, we can rearrange the equation E = hf to solve for f:
f = E / h
Plugging in the values, we get:
f = (2.095 × 10^-18 J) / (6.626 × 10^-34 J s) ≈ 3.160 × 10^15 Hz
Step 4: Find the wavelength of the photon.
The relationship between frequency (f) and wavelength (λ) is given by the equation c = fλ, where c is the speed of light (approximately 3.00 × 10^8 m/s). We can rearrange this equation to solve for λ:
λ = c / f
Plugging in the values, we get:
λ = (3.00 × 10^8 m/s) / (3.160 × 10^15 Hz) ≈ 9.49 × 10^-8 meters (or 94.9 nm)
Therefore, a photon of wavelength approximately 94.9 nm is required to ionize a hydrogen atom in the ground state and give the ejected electron a kinetic energy of 13.1 eV.