A photon with a wavelength (λ) of 3.091 x 10–7 m strikes an atom of hydrogen. Determine the velocity (in meters/second) of an electron ejected from the excited state, n = 3.
To determine the velocity of an electron ejected from an excited state of hydrogen, we can use the Rydberg formula and conservation of energy.
1. Calculate the energy of the incident photon:
Energy of a photon (E) = Planck's constant (h) * speed of light (c) / wavelength (λ)
E = (6.626 x 10^-34 J*s) * (3.0 x 10^8 m/s) / (3.091 x 10^-7 m)
2. Determine the energy difference between the initial and final states:
ΔE = E - E_n = E - (-13.6 eV / n^2)
3. Calculate the energy of the ejected electron:
ΔE = 13.6 eV / 3^2 - 13.6 eV / n^2
4. Convert the energy of the ejected electron to joules:
ΔE (J) = ΔE (eV) * 1.6 x 10^-19 J/eV
5. Use the energy to calculate the velocity of the ejected electron:
ΔE (J) = 1/2 * m * v^2, where m is the mass of the electron (9.109 x 10^-31 kg)
v = √(2 * ΔE / m)
Plug in the known values and calculate:
1. E = (6.626 x 10^-34 J*s) * (3.0 x 10^8 m/s) / (3.091 x 10^-7 m)
E ≈ 2.031 x 10^-19 J
2. ΔE = 2.031 x 10^-19 J - (-13.6 eV / 3^2) ≈ 2.031 x 10^-19 J + 1.51 x 10^-19 J ≈ 3.541 x 10^-19 J
3. ΔE (J) = 3.541 x 10^-19 J
4. ΔE (J) = 3.541 x 10^-19 J * 1.6 x 10^-19 J/eV ≈ 5.6656 J
5. v = √(2 * 5.6656 J / (9.109 x 10^-31 kg))
v ≈ 2.19 x 10^6 m/s
Therefore, the velocity of the ejected electron is approximately 2.19 x 10^6 m/s.
To determine the velocity of an electron ejected from the excited state in hydrogen, we can use the equations from the Bohr model of the hydrogen atom. The energy of a photon can be calculated using the equation:
E = hc / λ
Where:
E is the energy of the photon
h is Planck's constant (6.626 x 10^-34 J·s)
c is the speed of light (3.00 x 10^8 m/s)
λ is the wavelength of the photon
Once we know the energy of the photon, we can calculate the energy of the electron in the excited state using the equation:
En = -13.6 eV / n^2
Where:
En is the energy of the electron in the excited state
-13.6 eV is the ionization energy of hydrogen
n is the principal quantum number of the excited state
The velocity of the electron can be determined using the equation:
v = √(2(E - En) / me)
Where:
v is the velocity of the electron
E is the energy of the photon
En is the energy of the electron in the excited state
me is the mass of the electron (9.109 x 10^-31 kg)
Let's plug in the values to calculate the velocity:
E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (3.091 x 10^-7 m)
E = 6.81 x 10^-19 J
En = -13.6 eV / (3^2)
En = -1.511 eV
v = √(2(6.81 x 10^-19 J - (-1.511 eV * 1.602 x 10^-19 J/eV)) / (9.109 x 10^-31 kg))
v = √(2(6.81 x 10^-19 J - (-2.426 x 10^-19 J))/ (9.109 x 10^-31 kg))
v = √(2(9.236 x 10^-19 J) / (9.109 x 10^-31 kg))
v = √(1.845 x 10^12 m^2/s^2)
v = 4.30 x 10^6 m/s
Therefore, the velocity of the electron ejected from the excited state of hydrogen is approximately 4.30 x 10^6 m/s.