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A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :
- a)2
- b)4
- c)5
- d)3
Correct answer is option 'B'. Can you explain this answer?
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A hydrogen atom makes a transition from n = 2 to n = 1 and emits a ph...
Understanding the Hydrogen Transition
When a hydrogen atom transitions from n = 2 to n = 1, it emits a photon. The energy of this photon can be calculated using the formula:
- E = 13.6 eV * (1/n1^2 - 1/n2^2)
For n1 = 1 and n2 = 2:
- E = 13.6 eV * (1/1^2 - 1/2^2) = 13.6 eV * (1 - 1/4) = 10.2 eV
Photon Interaction with Doubly Ionized Lithium
The emitted photon then strikes a doubly ionized lithium atom (Li²⁺), which has an atomic number (Z) of 3. The energy required to remove the electron from an excited state is dependent on the quantum number of that state.
Energy Levels of Li²⁺
The energy levels for hydrogen-like atoms are given by the formula:
- E_n = -Z² * 13.6 eV / n²
For Li²⁺ (Z = 3), the energy levels become:
- E_n = -9.12 eV / n²
Calculating Minimum Quantum Number for Excitation
To find the minimum quantum number (n) for the excited state from which the electron can be removed by a 10.2 eV photon:
- Set E_n equal to the energy of the photon:
- 9.12 eV / n² = 10.2 eV
Solving for n:
- n² = 9.12 / 10.2
- n² ≈ 0.89
Since n must be a whole number, the lowest integer n must be 4 (since n = 3 would yield insufficient energy).
Conclusion
Thus, the least quantum number for the excited state of the doubly ionized lithium atom to allow the removal of the electron by the emitted photon from the hydrogen transition is:
- Option B: 4
When a hydrogen atom transitions from n = 2 to n = 1, it emits a photon. The energy of this photon can be calculated using the formula:
- E = 13.6 eV * (1/n1^2 - 1/n2^2)
For n1 = 1 and n2 = 2:
- E = 13.6 eV * (1/1^2 - 1/2^2) = 13.6 eV * (1 - 1/4) = 10.2 eV
Photon Interaction with Doubly Ionized Lithium
The emitted photon then strikes a doubly ionized lithium atom (Li²⁺), which has an atomic number (Z) of 3. The energy required to remove the electron from an excited state is dependent on the quantum number of that state.
Energy Levels of Li²⁺
The energy levels for hydrogen-like atoms are given by the formula:
- E_n = -Z² * 13.6 eV / n²
For Li²⁺ (Z = 3), the energy levels become:
- E_n = -9.12 eV / n²
Calculating Minimum Quantum Number for Excitation
To find the minimum quantum number (n) for the excited state from which the electron can be removed by a 10.2 eV photon:
- Set E_n equal to the energy of the photon:
- 9.12 eV / n² = 10.2 eV
Solving for n:
- n² = 9.12 / 10.2
- n² ≈ 0.89
Since n must be a whole number, the lowest integer n must be 4 (since n = 3 would yield insufficient energy).
Conclusion
Thus, the least quantum number for the excited state of the doubly ionized lithium atom to allow the removal of the electron by the emitted photon from the hydrogen transition is:
- Option B: 4
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A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :a)2b)4c)5d)3Correct answer is option 'B'. Can you explain this answer?
Question Description
A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :a)2b)4c)5d)3Correct answer is option 'B'. 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 A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :a)2b)4c)5d)3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :a)2b)4c)5d)3Correct answer is option 'B'. Can you explain this answer?.
A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :a)2b)4c)5d)3Correct answer is option 'B'. 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 A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :a)2b)4c)5d)3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :a)2b)4c)5d)3Correct answer is option 'B'. Can you explain this answer?.
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