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.
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To determine the velocity of the electron ejected from the excited state, we can use the equation for the energy of a photon:
E = hc/λ
Where:
E is the energy of the photon,
h is Planck's constant (6.626 x 10^-34 Js),
c is the speed of light (3.0 x 10^8 m/s), and
λ is the wavelength of the photon.
First, we need to calculate the energy of the photon:
E = (6.626 x 10^-34 Js) x (3.0 x 10^8 m/s) / (3.091 x 10^-7 m)
E ≈ 6.804 x 10^-19 J
Next, we can use the equation for the energy of an electron in a hydrogen atom:
E = -(13.6 eV) / n^2
Where:
E is the energy of the electron,
n is the principal quantum number (in this case, n=3), and
13.6 eV is the ionization energy of a hydrogen atom.
Using this equation, we can solve for the energy of the electron:
6.804 x 10^-19 J = -(13.6 eV) / (3^2)
Now, we convert this energy from electron volts (eV) to joules (J):
6.804 x 10^-19 J = -13.6 eV x 1.602 x 10^-19 J/eV / 9
Finally, we can use the equation for the kinetic energy of the electron to find its velocity:
E_kinetic = (1/2)mv^2
Where:
E_kinetic is the kinetic energy of the electron,
m is the mass of the electron (9.11 x 10^-31 kg), and
v is the velocity of the electron.
From our previous calculation, we know the energy of the electron (-6.804 x 10^-19 J). We can equate this to the kinetic energy:
-(6.804 x 10^-19 J) = (1/2)(9.11 x 10^-31 kg)v^2
Now we can solve for the velocity of the electron:
v^2 = -[(6.804 x 10^-19 J) x 2] / (9.11 x 10^-31 kg)
v^2 = -1.487 x 10^11 m^2/s^2
v ≈ ±1.219 x 10^5 m/s
The velocity of the electron ejected from the excited state is approximately ±1.219 x 10^5 m/s.
To determine the velocity of an electron ejected from the excited state of an atom, you can use the Rydberg formula and conservation of energy.
The Rydberg formula is given by:
1/λ = R * (1/n₁² - 1/n₂²),
where λ is the wavelength of the photon, R is the Rydberg constant (approximately 1.097 x 10^7 m^-1), n₁ is the initial energy level (ground state), and n₂ is the excited energy level.
In this case, we are given that the initial energy level (n₁) is the ground state, which corresponds to n = 1.
Let's plug in the values:
1/3.091 x 10–7 = R * (1/1² - 1/3²)
Simplifying the equation:
1/3.091 x 10–7 = R * (1 - 1/9)
1/3.091 x 10–7 = R * (8/9)
To determine R * (8/9), we need the value of the Rydberg constant, which is approximately 1.097 x 10^7 m^-1.
Let's calculate R * (8/9):
R * (8/9) ≈ 1.097 x 10^7 m^-1 * (8/9) ≈ 0.9722 x 10^7 m^-1
Now, we can rewrite the equation:
1/3.091 x 10–7 = 0.9722 x 10^7 m^-1
To solve for the wavelength λ, we need to take the reciprocal of both sides:
3.091 x 10–7 m = 1 / (0.9722 x 10^7 m^-1)
Simplifying the equation:
3.091 x 10–7 m = 1.0285 x 10^-8 m
Since the ejected electron is assumed to have negligible mass, we can use the equation to calculate its velocity:
v = c * (λ / λ₀),
where v is the velocity of the electron, c is the speed of light (approximately 3 x 10^8 m/s), λ is the wavelength of the photon, and λ₀ is the wavelength of the ejected electron.
In this case, λ is the wavelength of the photon (3.091 x 10–7 m) and λ₀ is the wavelength of the ejected electron, which we need to find.
Let's calculate the velocity of the electron:
v = 3 x 10^8 m/s * (3.091 x 10–7 m / 1.0285 x 10^-8 m)
Simplifying the equation:
v ≈ 9.1682 x 10^8 m/s
Therefore, the velocity of the electron ejected from the excited state, n = 3, is approximately 9.1682 x 10^8 m/s.