Atoms & Nuclei

JEE Main level, single correct, medium

69 of 239

Question 69 of 239

The energy required to ionise a hydrogen like ion in its ground state is 99 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited stale to the ground state?

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Answer: (B) 11.4nm11.4nm

hcλ=(13.6eV)z2[1n12−1n22]\dfrac{hc}{\lambda }=\left(13.6eV\right){z}^{2}\left[\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right]
n1=1{n}_{1}=1
n2=3{n}_{2}=3
hcλ=(13.6eV)(32)[112−132]\dfrac{hc}{\lambda }=\left(13.6eV\right)\left({3}^{2}\right)\left[\dfrac{1}{{1}^{2}}-\dfrac{1}{{3}^{2}}\right]
⇒hcλ=(13.6eV)(9)(89)\Rightarrow \dfrac{hc}{\lambda }=\left(13.6eV\right)\left(9\right)\left(\dfrac{8}{9}\right)
Wavelength =12408×13.6nm=\dfrac{1240}{8\times 13.6}\text{nm}
λ=11.39nm\lambda =11.39\text{nm}

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