Atoms & Nuclei

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13 of 239

Question 13 of 239

Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength λ\lambda. The value of principal quantum number nn of the excited state will be :
( R:R: Rydberg constant)

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Answer: (B) λRλR−1\sqrt{\dfrac{\lambda R}{\lambda R-1}}

When an electron jumps from a nth{n}^{th} orbit to ground state, using Rydberg formula,
1λ=R(1n12−1n22)\dfrac{1}{\lambda }=R\left(\dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}}\right)
In this case, n1=1{n}_{1}=1 and n2=n{n}_{2}=n. Therefore,
⇒1λ=R(112−1n2)⇒n=λRλR−1\Rightarrow \dfrac{1}{\lambda }=R\left(\dfrac{1}{{1}^{2}}-\dfrac{1}{{n}^{2}}\right) \Rightarrow n=\sqrt{\dfrac{\lambda R}{\lambda R-1}}

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