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A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980Å. The radius of the atom in the excited state, will be _____a0.
(hc = 12500 eV - Å)
(hc = 12500 eV - Å)
Correct answer is '16'. Can you explain this answer?
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A hydrogen atom, initially in the ground state is excited by absorbing...
Energy of photon = 12500/980 = 12.75 eV
∴ Electron will excite to n = 4
Since 'R' ∝ n2
∴ Radius of atom will be 16a0.
Most Upvoted Answer
A hydrogen atom, initially in the ground state is excited by absorbing...
To determine the final state of the hydrogen atom after absorbing a photon of wavelength 980 nm, we need to consider the energy levels of the hydrogen atom.
The energy levels of a hydrogen atom are given by the equation:
E = -13.6 eV / n^2
where E is the energy of the level, n is the principal quantum number, and -13.6 eV is the ionization energy of a hydrogen atom.
The energy of a photon is given by 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 (3.0 x 10^8 m/s), and λ is the wavelength of the photon.
First, let's convert the wavelength of the photon from nm to m:
980 nm = 980 x 10^-9 m
Now we can calculate the energy of the photon:
E = (6.626 x 10^-34 J*s)(3.0 x 10^8 m/s) / (980 x 10^-9 m)
E ≈ 2.014 x 10^-19 J
Next, we can equate the energy of the photon to the energy difference between the initial and final states of the hydrogen atom:
E = E_final - E_initial
Since the atom is initially in the ground state (n = 1), the initial energy is:
E_initial = -13.6 eV / (1^2) = -13.6 eV
Now we can solve for the final state:
E_final = E_initial + E
E_final = -13.6 eV + 2.014 x 10^-19 J
Finally, we can convert the final energy from eV to Joules:
1 eV = 1.602 x 10^-19 J
E_final = (-13.6 eV + 2.014 x 10^-19 J)(1.602 x 10^-19 J/eV)
E_final ≈ -21.824 eV
Therefore, the hydrogen atom is excited to the final state with an energy level of approximately -21.824 eV.
The energy levels of a hydrogen atom are given by the equation:
E = -13.6 eV / n^2
where E is the energy of the level, n is the principal quantum number, and -13.6 eV is the ionization energy of a hydrogen atom.
The energy of a photon is given by 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 (3.0 x 10^8 m/s), and λ is the wavelength of the photon.
First, let's convert the wavelength of the photon from nm to m:
980 nm = 980 x 10^-9 m
Now we can calculate the energy of the photon:
E = (6.626 x 10^-34 J*s)(3.0 x 10^8 m/s) / (980 x 10^-9 m)
E ≈ 2.014 x 10^-19 J
Next, we can equate the energy of the photon to the energy difference between the initial and final states of the hydrogen atom:
E = E_final - E_initial
Since the atom is initially in the ground state (n = 1), the initial energy is:
E_initial = -13.6 eV / (1^2) = -13.6 eV
Now we can solve for the final state:
E_final = E_initial + E
E_final = -13.6 eV + 2.014 x 10^-19 J
Finally, we can convert the final energy from eV to Joules:
1 eV = 1.602 x 10^-19 J
E_final = (-13.6 eV + 2.014 x 10^-19 J)(1.602 x 10^-19 J/eV)
E_final ≈ -21.824 eV
Therefore, the hydrogen atom is excited to the final state with an energy level of approximately -21.824 eV.
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A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980Å. The radius of the atom in the excited state, will be _____a0.(hc= 12500 eV - Å)Correct answer is '16'. Can you explain this answer?
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