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Ionization potential of hydrogen atom is 13.6eV.Hydrogen atoms in the ground state are excitedby monochromatic radiation of photon energy12.1 eV. According to Bohr’s theory, the spectrallines emitted by hydrogen will be [2006]
  • a)
    three
  • b)
    Four
  • c)
    One
  • d)
    Two
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Ionization potential of hydrogen atom is 13.6eV.Hydrogen atoms in the ...
Energy of ground state 13.6 eV
Energy of first excited state
Energy of second excited state
Difference between ground state and 2nd
excited state = 13.6 – 1.5 = 12.1 eV
So, electron can be excited upto 3rd orbit
No. of possible transition
1 → 2, 1 → 3, 2 → 3
So, three lines are possible
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Most Upvoted Answer
Ionization potential of hydrogen atom is 13.6eV.Hydrogen atoms in the ...
's theory, what will happen to the hydrogen atom after absorbing the photon?

According to Bohr's theory, when the hydrogen atom absorbs a photon of energy equal to or greater than its ionization potential (13.6 eV), the electron in the ground state will be excited to a higher energy level or even completely ionized.

In this case, the photon energy (12.1 eV) is less than the ionization potential, but still enough to excite the electron to a higher energy level. The electron will move to a higher energy level, and the hydrogen atom will be in an excited state.

The electron will eventually return to the ground state, releasing the excess energy as a photon of a specific wavelength. The wavelength of the emitted photon can be calculated using the Rydberg formula:

1/λ = RZ^2(1/nf^2 - 1/ni^2)

where λ is the wavelength of the emitted photon, R is the Rydberg constant (1.0974 x 10^7 m^-1), Z is the atomic number (1 for hydrogen), and nf and ni are the final and initial energy levels, respectively.

Assuming the electron returns to the ground state (nf = 1), the emitted photon will have a wavelength of:

1/λ = R(1/1^2 - 1/n^2)

where n is the initial energy level that the electron was excited to.

For n = 2 (the first excited state), the wavelength of the emitted photon will be:

1/λ = R(1 - 1/4) = 3R/4

λ = 4/3R = 97.1 nm

Therefore, after absorbing a photon of energy 12.1 eV, the hydrogen atom will be in an excited state, and will eventually emit a photon of wavelength 97.1 nm as the electron returns to the ground state.
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