calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n=7 to n=1. Is it just 6? or how do I do this?
2.136 x 10^-18
To calculate the energy of a photon emitted during a transition of an electron in a hydrogen atom, you can use the following formula:
E = -13.6 × (Z^2 / n^2) electron volts (eV)
Where:
- E is the energy of the photon in eV.
- Z is the atomic number, which is 1 for hydrogen.
- n is the principal quantum number.
In this case, the electron is transitioning from n = 7 to n = 1. Plugging these values into the formula, we have:
E = -13.6 × (1^2 / 1^2) eV
E = -13.6 eV
Therefore, the energy of the photon emitted during this transition is -13.6 electron volts (eV). Note that the negative sign indicates that energy is released during the transition.
To calculate the energy of a photon emitted during an electron transition, you can use the equation:
E = hf
where E is the energy, h is Planck's constant (approx. 6.63 x 10^-34 J.s), and f is the frequency of the photon.
First, you need to determine the frequency of the emitted photon. The frequency can be calculated using the Rydberg formula:
1/λ = R[(1/n1^2) - (1/n2^2)]
where λ is the wavelength, R is the Rydberg constant (approx. 1.097 x 10^7 m^-1), n1 and n2 are the initial and final energy levels of the electron transition, respectively.
In this case, the transition is from n = 7 to n = 1. Substituting these values into the Rydberg formula:
1/λ = R[(1/1^2) - (1/7^2)]
Simplifying this expression:
1/λ = R(1 - 1/49)
1/λ = R(48/49)
Now, we can rearrange this equation to solve for the wavelength, λ:
λ = 49/(48R)
Next, we can use the speed of light, c (approx. 3.00 x 10^8 m/s), to calculate the frequency:
f = c/λ
Substituting the value of λ we calculated earlier:
f = c/(49/(48R))
Simplifying this expression:
f = (48Rc)/49
Finally, we can substitute the value of the frequency into the energy equation:
E = hf = (hf = (48Rc)/49) x h
Calculating this expression gives you the energy of the photon emitted during the electron transition from n = 7 to n = 1 in a hydrogen atom.
No.
delta E = 2.180 x 10^-18 J x (1/n1^2 - 1/n2^2)
n1 = 1 and 1^2 = 1
n2 = 7 and 7^2 = 49
Solve for delta E.