The ionization energy of the electron in thehydrogen atom in its ground state is 13.6 eV.The atoms are excited to higher energy levels toemit radiations of 6 wavelengths. Maximumwavelength of emitted radiation corresponds tothe transition between [2009]a)n = 3 to n = 1 statesb)n = 2 to n = 1 statesc)n = 4 to n = 3 statesd)n = 3 to n = 2 statesCorrect answer is option 'C'. Can you explain this answer?

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a)
n = 3 to n = 1 states
b)
n = 2 to n = 1 states
c)
n = 4 to n = 3 states
d)
n = 3 to n = 2 states

Correct answer is option 'C'. Can you explain this answer?

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The ionization energy of the electron in thehydrogen atom in its groun...

n² – n – 12 = 0
(n – 4) (n + 3) = 0 or n = 4

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The ionization energy of the electron in thehydrogen atom in its groun...

Explanation:
When hydrogen atoms are excited to higher energy levels, they release energy in the form of photons. 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, c is the speed of light, and λ is the wavelength of the photon.

The maximum wavelength of the emitted radiation corresponds to the transition that releases the least amount of energy. This occurs when the electron transitions from a higher energy level to the ground state.

To determine which transition corresponds to the maximum wavelength of the emitted radiation, we can use the equation:

ΔE = -Rh/n^2f + Rh/n^2i

where ΔE is the energy released during the transition, Rh is the Rydberg constant (2.18 x 10^-18 J), nif is the initial energy level, and nf is the final energy level.

We want to find the transition that releases the least amount of energy, which corresponds to the maximum wavelength of the emitted radiation. This occurs when nf is as large as possible and nif is as small as possible.

Using this approach, we find that the transition between n = 4 and n = 3 states releases the least amount of energy, and therefore corresponds to the maximum wavelength of the emitted radiation. The correct answer is option C.

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The ionization energy of the electron in thehydrogen atom in its ground state is 13.6 eV.The atoms are excited to higher energy levels toemit radiations of 6 wavelengths. Maximumwavelength of emitted radiation corresponds tothe transition between [2009]a)n = 3 to n = 1 statesb)n = 2 to n = 1 statesc)n = 4 to n = 3 statesd)n = 3 to n = 2 statesCorrect answer is option 'C'. Can you explain this answer?

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The ionization energy of the electron in thehydrogen atom in its ground state is 13.6 eV.The atoms are excited to higher energy levels toemit radiations of 6 wavelengths. Maximumwavelength of emitted radiation corresponds tothe transition between [2009]a)n = 3 to n = 1 statesb)n = 2 to n = 1 statesc)n = 4 to n = 3 statesd)n = 3 to n = 2 statesCorrect answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The ionization energy of the electron in thehydrogen atom in its ground state is 13.6 eV.The atoms are excited to higher energy levels toemit radiations of 6 wavelengths. Maximumwavelength of emitted radiation corresponds tothe transition between [2009]a)n = 3 to n = 1 statesb)n = 2 to n = 1 statesc)n = 4 to n = 3 statesd)n = 3 to n = 2 statesCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ionization energy of the electron in thehydrogen atom in its ground state is 13.6 eV.The atoms are excited to higher energy levels toemit radiations of 6 wavelengths. Maximumwavelength of emitted radiation corresponds tothe transition between [2009]a)n = 3 to n = 1 statesb)n = 2 to n = 1 statesc)n = 4 to n = 3 statesd)n = 3 to n = 2 statesCorrect answer is option 'C'. Can you explain this answer?.

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