The electron in a ground state H atom absorbs a photon of wavelength 121.57nm, to what energy level does the electron move?
1/wavelength = R(1/1^2 - 1/n^2)
Remember to use wavelength in m.
R is 1.0973E7
The 1/1^2 in the above is 1/n1 and n1 is squared.
The 1/n^2 in the above is 1/n2 and the n2 is squared.
Solve for n.
This 121.6 nm line is the first line of the Lyman series and has a value for n of 2.
Calculate the energy difference for the transition of n=6 to n=1 for 1.00 mol of hydrogen atom?
To determine the energy level to which the electron moves when absorbing a photon of wavelength 121.57 nm, we can use the equation:
ΔE = hc/λ,
where ΔE is the change in energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon.
First, we need to convert the wavelength from nm to m:
λ = 121.57 nm × (1 m / 10^9 nm) = 1.2157 x 10^-7 m.
Now we can calculate ΔE:
ΔE = (6.626 x 10^-34 J·s) × (2.998 x 10^8 m/s) / (1.2157 x 10^-7 m).
Calculating ΔE, we find:
ΔE = 1.6323 x 10^-18 J.
The energy change ΔE corresponds to the transition from one energy level to another. In the hydrogen atom, the energy levels are described by the formula:
En = -13.6 eV / n^2,
where En is the energy of the nth energy level measured in electron volts (eV).
Let's solve for n:
1.6323 x 10^-18 J = (-13.6 eV / n^2) × (1.602 x 10^-19 J/eV).
Simplifying the equation and rearranging, we have:
n^2 = (-13.6 eV / (1.6323 x 10^-18 J)) × (1.602 x 10^-19 J/eV).
n^2 ≈ 13.026.
Taking the square root of both sides:
n ≈ √13.026,
n ≈ 3.61.
Therefore, the electron moves to the 4th energy level (n = 4) when it absorbs a photon of wavelength 121.57 nm.
To determine the energy level that the electron moves to after absorbing a photon, we need to use the equation relating the energy of a photon to its wavelength.
The energy (E) of a photon is given by the equation:
E = hc/λ,
where h is the Planck constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the photon.
Let's calculate the energy of the photon using the given wavelength of 121.57 nm:
First, convert the wavelength to meters:
λ = 121.57 nm = 121.57 x 10^-9 m.
Now, substitute the values into the equation:
E = (6.626 x 10^-34 J·s) * (3.00 x 10^8 m/s) / (121.57 x 10^-9 m).
Calculating the equation gives:
E ≈ 5.11 x 10^-19 J.
Next, we need to find which energy level this corresponds to in a hydrogen atom.
The energy levels in a hydrogen atom are given by the equation:
En = -13.6 eV/n^2,
where En is the energy level, -13.6 eV is the ionization energy of hydrogen, and n is the principal quantum number representing the energy level.
To find the energy level (n), we can rearrange the equation by solving for n:
n^2 = -13.6 eV / En.
n^2 = -13.6 eV / (5.11 x 10^-19 J × (1 eV / 1.6 x 10^-19 J)).
Simplifying the equation gives:
n^2 ≈ 16.
Therefore, n ≈ 4.
Hence, the electron moves to the 4th energy level (n = 4) in the hydrogen atom after absorbing the photon.