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Which hydrogenic atom ion has the wavelength difference between first line of Balmer and Lymen Series equal to 59.3 nm? (RH = 109678 cm–1).
- a)Z = 3
- b)Z = 2
- c)Z = 4
- d)Z = 1
Correct answer is option 'A'. Can you explain this answer?
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To find the hydrogenic atom ion that has a wavelength difference between the first line of the Balmer and Lyman series equal to 59.3 nm, we can use the Rydberg formula:
1/λ = RH * (1/n1^2 - 1/n2^2)
where λ is the wavelength, RH is the Rydberg constant (109678 cm^-1), and n1 and n2 are the principal quantum numbers of the energy levels involved.
For the Balmer series, n1 = 2 and n2 = 3. Plugging these values into the formula, we have:
1/λ_balmer = 109678 * (1/2^2 - 1/3^2)
1/λ_balmer = 109678 * (1/4 - 1/9)
1/λ_balmer = 109678 * (9/36 - 4/36)
1/λ_balmer = 109678 * (5/36)
1/λ_balmer = 1524.972
For the Lyman series, n1 = 1 and n2 = 2. Plugging these values into the formula, we have:
1/λ_lyman = 109678 * (1/1^2 - 1/2^2)
1/λ_lyman = 109678 * (1/1 - 1/4)
1/λ_lyman = 109678 * (3/4)
1/λ_lyman = 82108.5
The difference in wavelength between the first line of the Balmer and Lyman series is:
Δλ = λ_lyman - λ_balmer
Δλ = 82108.5 - 1524.972
Δλ = 80583.528
Therefore, there is no hydrogenic atom ion that has a wavelength difference between the first line of the Balmer and Lyman series equal to 59.3 nm.
1/λ = RH * (1/n1^2 - 1/n2^2)
where λ is the wavelength, RH is the Rydberg constant (109678 cm^-1), and n1 and n2 are the principal quantum numbers of the energy levels involved.
For the Balmer series, n1 = 2 and n2 = 3. Plugging these values into the formula, we have:
1/λ_balmer = 109678 * (1/2^2 - 1/3^2)
1/λ_balmer = 109678 * (1/4 - 1/9)
1/λ_balmer = 109678 * (9/36 - 4/36)
1/λ_balmer = 109678 * (5/36)
1/λ_balmer = 1524.972
For the Lyman series, n1 = 1 and n2 = 2. Plugging these values into the formula, we have:
1/λ_lyman = 109678 * (1/1^2 - 1/2^2)
1/λ_lyman = 109678 * (1/1 - 1/4)
1/λ_lyman = 109678 * (3/4)
1/λ_lyman = 82108.5
The difference in wavelength between the first line of the Balmer and Lyman series is:
Δλ = λ_lyman - λ_balmer
Δλ = 82108.5 - 1524.972
Δλ = 80583.528
Therefore, there is no hydrogenic atom ion that has a wavelength difference between the first line of the Balmer and Lyman series equal to 59.3 nm.
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Which hydrogenic atom ion has the wavelength difference between first line of Balmer and Lymen Series equal to 59.3 nm? (RH = 109678 cm–1).a)Z = 3b)Z = 2c)Z = 4d)Z = 1Correct answer is option 'A'. Can you explain this answer?
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