= Find the energy of the photon released in the transition from n₁ for a hydrogen atom. (Note: Use Rydberg Formula) 3 to n₂ = 2
Q: Given the value of Rydberg constant is 107m−1, the waves number of the lest line of the Balmer…
A: We know that: λ1= R(1/n12−1/n22)
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A: Energy of transition for an atom is given by: ΔE = R(1/nf2 - 1/ni2) where, ΔE = energy released R=…
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A: we have n2 = 2 λ=434nm = 434×10-9m RH = 1.09737316×107 m-1
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A: Here, 1λ=R1n12-1n22
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A: Given, Rydberg's constant, R=1.0973x107 m-1 n1=6 n2=12
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Q: Find the wavelengths of the transitions the transition from n₁ = 7 to n₂ = 2 for a hydrogen atom.
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A: Solution: The energy corresponding to the given wavelength is calculated as; E=1241486.1 =2.55×10-9…
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A: Given that, n2= 1 n1 = 7 Rydberg constant R= 2.18×10-18 J
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A: The wavelength is, E=hcλ⇒λ=hcE
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A: Energy of the photon, E = 1.633 x 10-18 J
Q: For what energy levels in the hydrogen atom will we not find l = 2 states?
A: To determine:For what energy levels in the hydrogen atom will we not find l = 2 states?
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