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Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. The spectral lines emitted by hydrogen atoms according to Bohr's theory will be
- a)One
- b)Two
- c)Three
- d)Four
Correct answer is option 'C'. Can you explain this answer?
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Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in t...
Explanation:
Bohr's Theory of Hydrogen Atom states that when an electron jumps from a higher energy level to a lower energy level, it emits a photon of specific frequency/wavelength. The frequency of the photon emitted is given by the equation:
ΔE = hf
where ΔE is the energy difference between the higher and lower energy levels, h is Planck's constant (6.626 x 10^-34 J s), and f is the frequency of the emitted photon.
In the given problem, the hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. This means that the electron in the hydrogen atom absorbs a photon of energy 12.1 eV and jumps to a higher energy level. The energy difference between the ground state and the excited state is given by:
ΔE = 12.1 eV - 13.6 eV = -1.5 eV
The negative sign indicates that energy is absorbed by the electron. Now, the electron will eventually return to the ground state by emitting photons of specific frequencies/wavelengths.
The spectral lines emitted by hydrogen atoms according to Bohr's theory can be calculated using the equation:
1/λ = RZ^2(1/n1^2 - 1/n2^2)
where λ is the wavelength of the emitted photon, R is the Rydberg constant (1.097 x 10^7 m^-1), Z is the atomic number (1 for hydrogen), n1 is the initial energy level, and n2 is the final energy level.
In this case, the electron is returning from an excited state to the ground state. Therefore, n1 = 2 and n2 = 1. Substituting these values in the above equation, we get:
1/λ = R(1/1^2 - 1/2^2) = 3R/4
λ = 4/3R ≈ 656.3 nm
This corresponds to the red spectral line in the Balmer series of hydrogen atom. Similarly, the electron can jump from the excited state to the 2nd energy level (n1 = 2, n2 = 2), 3rd energy level (n1 = 2, n2 = 3), or 4th energy level (n1 = 2, n2 = 4) before returning to the ground state. Therefore, a total of three spectral lines will be emitted by the hydrogen atoms according to Bohr's theory.
Bohr's Theory of Hydrogen Atom states that when an electron jumps from a higher energy level to a lower energy level, it emits a photon of specific frequency/wavelength. The frequency of the photon emitted is given by the equation:
ΔE = hf
where ΔE is the energy difference between the higher and lower energy levels, h is Planck's constant (6.626 x 10^-34 J s), and f is the frequency of the emitted photon.
In the given problem, the hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. This means that the electron in the hydrogen atom absorbs a photon of energy 12.1 eV and jumps to a higher energy level. The energy difference between the ground state and the excited state is given by:
ΔE = 12.1 eV - 13.6 eV = -1.5 eV
The negative sign indicates that energy is absorbed by the electron. Now, the electron will eventually return to the ground state by emitting photons of specific frequencies/wavelengths.
The spectral lines emitted by hydrogen atoms according to Bohr's theory can be calculated using the equation:
1/λ = RZ^2(1/n1^2 - 1/n2^2)
where λ is the wavelength of the emitted photon, R is the Rydberg constant (1.097 x 10^7 m^-1), Z is the atomic number (1 for hydrogen), n1 is the initial energy level, and n2 is the final energy level.
In this case, the electron is returning from an excited state to the ground state. Therefore, n1 = 2 and n2 = 1. Substituting these values in the above equation, we get:
1/λ = R(1/1^2 - 1/2^2) = 3R/4
λ = 4/3R ≈ 656.3 nm
This corresponds to the red spectral line in the Balmer series of hydrogen atom. Similarly, the electron can jump from the excited state to the 2nd energy level (n1 = 2, n2 = 2), 3rd energy level (n1 = 2, n2 = 3), or 4th energy level (n1 = 2, n2 = 4) before returning to the ground state. Therefore, a total of three spectral lines will be emitted by the hydrogen atoms according to Bohr's theory.
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Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in t...
Energy in excited state = −13.6 + 12.1 = −1.5 eV
Number of spectral lines
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Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. The spectral lines emitted by hydrogen atoms according to Bohr's theory will bea)Oneb)Twoc)Threed)FourCorrect answer is option 'C'. Can you explain this answer?
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Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. The spectral lines emitted by hydrogen atoms according to Bohr's theory will bea)Oneb)Twoc)Threed)FourCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. The spectral lines emitted by hydrogen atoms according to Bohr's theory will bea)Oneb)Twoc)Threed)FourCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ionisation potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. The spectral lines emitted by hydrogen atoms according to Bohr's theory will bea)Oneb)Twoc)Threed)FourCorrect answer is option 'C'. Can you explain this answer?.
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