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At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?
Correct answer is '2'. Can you explain this answer?
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At what minimum atomic number, a transition from n = 2 to n = 1 energy...
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At what minimum atomic number, a transition from n = 2 to n = 1 energy...
Explanation:
The energy emitted during a transition from higher energy level to lower energy level can be calculated using the formula:
E = (hc)/λ
where E is the energy emitted, h is Planck's constant, c is the speed of light, and λ is the wavelength of the emitted radiation.
When an electron makes a transition from n=2 to n=1 energy level, the emitted radiation is in the form of X-rays. The minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m can be calculated as follows:
E = (hc)/λ
E = (6.626 x 10^-34 J·s) x (3.0 x 10^8 m/s) / (3.0 x 10^-8 m)
E = 6.626 x 10^-16 J
The energy difference between n=2 and n=1 energy levels can be calculated using the formula:
ΔE = (-13.6 eV) x [(1/nf^2) - (1/ni^2)]
where ΔE is the energy difference, nf is the final energy level (n=1), and ni is the initial energy level (n=2).
ΔE = (-13.6 eV) x [(1/1^2) - (1/2^2)]
ΔE = -10.2 eV
Converting this energy to joules:
1 eV = 1.602 x 10^-19 J
-10.2 eV = -1.64 x 10^-18 J
The minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m can be calculated using the formula:
ΔE = (Z^2 x 13.6 eV) / n^2
where Z is the atomic number and n is the initial energy level (n=2).
-1.64 x 10^-18 J = (Z^2 x 13.6 eV) / 2^2
Z^2 = (-1.64 x 10^-18 J x 4) / 13.6 eV
Z^2 = -4.8 x 10^-19
Z = 2
Therefore, the minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m is 2.
The energy emitted during a transition from higher energy level to lower energy level can be calculated using the formula:
E = (hc)/λ
where E is the energy emitted, h is Planck's constant, c is the speed of light, and λ is the wavelength of the emitted radiation.
When an electron makes a transition from n=2 to n=1 energy level, the emitted radiation is in the form of X-rays. The minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m can be calculated as follows:
E = (hc)/λ
E = (6.626 x 10^-34 J·s) x (3.0 x 10^8 m/s) / (3.0 x 10^-8 m)
E = 6.626 x 10^-16 J
The energy difference between n=2 and n=1 energy levels can be calculated using the formula:
ΔE = (-13.6 eV) x [(1/nf^2) - (1/ni^2)]
where ΔE is the energy difference, nf is the final energy level (n=1), and ni is the initial energy level (n=2).
ΔE = (-13.6 eV) x [(1/1^2) - (1/2^2)]
ΔE = -10.2 eV
Converting this energy to joules:
1 eV = 1.602 x 10^-19 J
-10.2 eV = -1.64 x 10^-18 J
The minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m can be calculated using the formula:
ΔE = (Z^2 x 13.6 eV) / n^2
where Z is the atomic number and n is the initial energy level (n=2).
-1.64 x 10^-18 J = (Z^2 x 13.6 eV) / 2^2
Z^2 = (-1.64 x 10^-18 J x 4) / 13.6 eV
Z^2 = -4.8 x 10^-19
Z = 2
Therefore, the minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m is 2.
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At what minimum atomic number, a transition from n = 2 to n = 1 energy...
∆E = RhcZ2 (1/n12 – 1/n22) Here, R = 1.0967 * 107 m-1 h = 6.626 * 10-34 J sec, c = 3 * 108 m/sec n1 = 1, n2 = 2 and for H-atom, Z = 1 E2 – E1 = 1.0967 * 107 * 6.626 * 10-34 * 3 * 108(1/1 – 1/4) ∆E = 1.0967 * 6.626 * 3 * ¾ * 10-19 J = 16.3512 * 10-19 J = 16.3512 *10-19/1.6 *10-19 eV = 10.22 eV ∆E = hc/λ = RhcZ2 (1/n12 -1/n22) 1/λ = RZ2 (1/1 – 1/4) = RZ2 * 3/4 Given, λ = 3 * 10-8= m ∴ 1/3 *10-8 = 1.0967 = Z2 * 3/4 * 107 ∴ Z2 = 108 *4/3 *3 *1.0967 *107 = 40/9 *1.0967 = 4 ∴ Z = 2
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At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?Correct answer is '2'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?Correct answer is '2'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?Correct answer is '2'. Can you explain this answer?.
At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?Correct answer is '2'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?Correct answer is '2'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?Correct answer is '2'. Can you explain this answer?.
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