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Physics · 2019

JEE Main · 12 April 2019, Shift 1 · Q28

An excited He^+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum…

An excited $\displaystyle \mathrm{He}^{+}$ ion emits two photons in succession, with wavelengths $\displaystyle 108.5$ nm and $\displaystyle 30.4$ nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength $\displaystyle \lambda$, energy $$\mathrm{E}=\frac{1240 \mathrm{eV}}{\lambda(\text { in } \mathrm{nm})}): $$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.