Atoms & Nuclei

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71 of 239

Question 71 of 239

An excited He+{He}^{+} ion emits two photons in succession, with wavelengths 108.5nm108.5nm and 30.4nm30.4nm in making a transition to the ground state. The quantum number nn, corresponding to its initial excited state is

(for a photon of wavelength λ\lambda, energy E=1240eVλ(innm)E=\dfrac{1240eV}{\lambda (innm)})

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Answer: (B) n=5n=5

E1=1240λ1=1240108.5≈11.43eV{E}_{1}=\dfrac{1240}{{\lambda }_{1}}=\dfrac{1240}{108.5}\approx 11.43eV
E2=1240λ2=124030.4≈40.79eV{E}_{2}=\dfrac{1240}{{\lambda }_{2}}=\dfrac{1240}{30.4}\approx 40.79eV
ETotal=E1+E1=52.22eV{E}_{Total}={E}_{1}+{E}_{1}=52.22eV
52.22=13.6(2)2[1−1n2]52.22=13.6{\left(2\right)}^{2}\left[1-\dfrac{1}{{n}^{2}}\right]
52.22=54.4(1−1n2)52.22=54.4\left(1-\dfrac{1}{{n}^{2}}\right)
0.96=1−1n20.96=1-\dfrac{1}{{n}^{2}}
1n2=0.04\dfrac{1}{{n}^{2}}=0.04
n2=10.04=1004=25{n}^{2}=\dfrac{1}{0.04}=\dfrac{100}{4}=25
n=5n=5

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