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A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____.(upto two decimal places)
Correct answer is '10.58'. Can you explain this answer?
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A hydrogen like atom of atomic number z is in an excited state of quan...
Let ground state energy (in eV) be E1,
Then from the given condition
From equation
E1 = -(13.6 )Z2 ⇒ z = 4
Emin = 10.58 eV
Then from the given condition
From equation
E1 = -(13.6 )Z2 ⇒ z = 4
Emin = 10.58 eVMost Upvoted Answer
A hydrogen like atom of atomic number z is in an excited state of quan...
Energy Emitted during Transition
When a hydrogen-like atom transitions from one quantum state to another, it emits a photon with energy equal to the difference in energy between the two states. This can be calculated using the formula:
E = 13.6 * (Z^2 / n^2) eV
Where:
- E is the energy of the emitted photon
- Z is the atomic number of the atom
- n is the principal quantum number of the initial state
Finding the Quantum State
Given that the atom is in an excited state with quantum number 2n, we can determine the principal quantum number of the initial state by dividing the given quantum number by 2:
n = (2n) / 2 = n
So, the initial state quantum number is also n.
Calculating the Energy of the Maximum Photon
The given maximum energy of the emitted photon is 204 eV. We can use this information to calculate the atomic number of the atom using the formula:
204 = 13.6 * (Z^2 / n^2)
Rearranging the equation, we have:
(Z^2 / n^2) = 204 / 13.6
(Z^2 / n^2) = 15
Z^2 = 15 * n^2
Finding the Energy of the Transition to State n
The energy of the photon emitted during the transition to state n is given as 40.8 eV. Using the formula mentioned earlier:
40.8 = 13.6 * (Z^2 / n^2)
Rearranging the equation, we have:
(Z^2 / n^2) = 40.8 / 13.6
(Z^2 / n^2) = 3
Z^2 = 3 * n^2
Finding the Minimum Energy Emitted
To find the minimum energy that can be emitted during the de-excitation of the atom, we need to find the difference in energy between the maximum and minimum transitions. Subtracting the equation for the minimum energy from the equation for the maximum energy:
15 * n^2 - 3 * n^2 = 204 - 40.8
12 * n^2 = 163.2
n^2 = 163.2 / 12
n^2 = 13.6
n = √13.6
n ≈ 3.68 (rounded to two decimal places)
The minimum energy emitted during de-excitation is given by:
E = 13.6 * (Z^2 / n^2)
E = 13.6 * (15 / 3.68^2)
E ≈ 10 eV
Therefore, the minimum energy that can be emitted by this atom during de-excitation is approximately 10 eV.
When a hydrogen-like atom transitions from one quantum state to another, it emits a photon with energy equal to the difference in energy between the two states. This can be calculated using the formula:
E = 13.6 * (Z^2 / n^2) eV
Where:
- E is the energy of the emitted photon
- Z is the atomic number of the atom
- n is the principal quantum number of the initial state
Finding the Quantum State
Given that the atom is in an excited state with quantum number 2n, we can determine the principal quantum number of the initial state by dividing the given quantum number by 2:
n = (2n) / 2 = n
So, the initial state quantum number is also n.
Calculating the Energy of the Maximum Photon
The given maximum energy of the emitted photon is 204 eV. We can use this information to calculate the atomic number of the atom using the formula:
204 = 13.6 * (Z^2 / n^2)
Rearranging the equation, we have:
(Z^2 / n^2) = 204 / 13.6
(Z^2 / n^2) = 15
Z^2 = 15 * n^2
Finding the Energy of the Transition to State n
The energy of the photon emitted during the transition to state n is given as 40.8 eV. Using the formula mentioned earlier:
40.8 = 13.6 * (Z^2 / n^2)
Rearranging the equation, we have:
(Z^2 / n^2) = 40.8 / 13.6
(Z^2 / n^2) = 3
Z^2 = 3 * n^2
Finding the Minimum Energy Emitted
To find the minimum energy that can be emitted during the de-excitation of the atom, we need to find the difference in energy between the maximum and minimum transitions. Subtracting the equation for the minimum energy from the equation for the maximum energy:
15 * n^2 - 3 * n^2 = 204 - 40.8
12 * n^2 = 163.2
n^2 = 163.2 / 12
n^2 = 13.6
n = √13.6
n ≈ 3.68 (rounded to two decimal places)
The minimum energy emitted during de-excitation is given by:
E = 13.6 * (Z^2 / n^2)
E = 13.6 * (15 / 3.68^2)
E ≈ 10 eV
Therefore, the minimum energy that can be emitted by this atom during de-excitation is approximately 10 eV.
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A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____.(upto two decimal places)Correct answer is '10.58'. Can you explain this answer?
Question Description
A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____.(upto two decimal places)Correct answer is '10.58'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____.(upto two decimal places)Correct answer is '10.58'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____.(upto two decimal places)Correct answer is '10.58'. Can you explain this answer?.
A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____.(upto two decimal places)Correct answer is '10.58'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____.(upto two decimal places)Correct answer is '10.58'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrogen like atom of atomic number z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 is emitted. The minimum energy (in eV) that can be emitted by this atom during de - excitation i s _____.(upto two decimal places)Correct answer is '10.58'. Can you explain this answer?.
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