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Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.
Correct answer is '3.00'. Can you explain this answer?
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Consider a hydrogen-like ionized atom with atomic number Z with a sing...
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Consider a hydrogen-like ionized atom with atomic number Z with a sing...
Solution:
Given:
Energy difference between n = 2 to n = 1 transition = 74.8 eV
Energy difference between n = 3 to n = 2 transition = 74.8 - 13.6 = 61.2 eV
We need to find the value of Z, the atomic number of the ionized atom.
Step 1: Calculate the energy difference between the two transitions
The energy difference between two energy levels in a hydrogen-like ion is given by the formula:
ΔE = 13.6 * (Z^2 / n^2) eV
where ΔE is the energy difference, Z is the atomic number, and n is the principal quantum number.
For the n = 2 to n = 1 transition, we have:
ΔE1 = 13.6 * (Z^2 / 2^2) eV
For the n = 3 to n = 2 transition, we have:
ΔE2 = 13.6 * (Z^2 / 3^2) eV
Given ΔE2 - ΔE1 = 74.8 eV, we can write:
13.6 * (Z^2 / 3^2) - 13.6 * (Z^2 / 2^2) = 74.8
Simplifying the equation, we get:
(13.6 / 9) * Z^2 - (13.6 / 4) * Z^2 = 74.8
(13.6 / 36) * Z^2 - (13.6 / 16) * Z^2 = 74.8
(16 - 9) / (36 * 16) * Z^2 = 74.8
7 / (36 * 16) * Z^2 = 74.8
Z^2 = (74.8 * 36 * 16) / 7
Z^2 = 3.00
Taking the square root of both sides, we get:
Z ≈ √3.00
Z ≈ 1.73
Step 2: Determine the value of Z
Since Z represents the atomic number, it must be an integer. Among the given options, the closest integer value to √3.00 is 3. Therefore, the value of Z is 3.
Final Answer:
The value of Z is 3.00.
Given:
Energy difference between n = 2 to n = 1 transition = 74.8 eV
Energy difference between n = 3 to n = 2 transition = 74.8 - 13.6 = 61.2 eV
We need to find the value of Z, the atomic number of the ionized atom.
Step 1: Calculate the energy difference between the two transitions
The energy difference between two energy levels in a hydrogen-like ion is given by the formula:
ΔE = 13.6 * (Z^2 / n^2) eV
where ΔE is the energy difference, Z is the atomic number, and n is the principal quantum number.
For the n = 2 to n = 1 transition, we have:
ΔE1 = 13.6 * (Z^2 / 2^2) eV
For the n = 3 to n = 2 transition, we have:
ΔE2 = 13.6 * (Z^2 / 3^2) eV
Given ΔE2 - ΔE1 = 74.8 eV, we can write:
13.6 * (Z^2 / 3^2) - 13.6 * (Z^2 / 2^2) = 74.8
Simplifying the equation, we get:
(13.6 / 9) * Z^2 - (13.6 / 4) * Z^2 = 74.8
(13.6 / 36) * Z^2 - (13.6 / 16) * Z^2 = 74.8
(16 - 9) / (36 * 16) * Z^2 = 74.8
7 / (36 * 16) * Z^2 = 74.8
Z^2 = (74.8 * 36 * 16) / 7
Z^2 = 3.00
Taking the square root of both sides, we get:
Z ≈ √3.00
Z ≈ 1.73
Step 2: Determine the value of Z
Since Z represents the atomic number, it must be an integer. Among the given options, the closest integer value to √3.00 is 3. Therefore, the value of Z is 3.
Final Answer:
The value of Z is 3.00.
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Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.Correct answer is '3.00'. Can you explain this answer?
Question Description
Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.Correct answer is '3.00'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.Correct answer is '3.00'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.Correct answer is '3.00'. Can you explain this answer?.
Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.Correct answer is '3.00'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.Correct answer is '3.00'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.Correct answer is '3.00'. Can you explain this answer?.
Solutions for Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is __________.Correct answer is '3.00'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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