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QJEEM19Ph3677: JEE Main 2019 Physics - Class 12 JEEM19Ph3677 Solution: An excited He + ion emits two photons in succession, with…

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 number n, corresponding to its initial excited state is (for photon of wavelength λ, energy E=1240eVλ(innm)) :

Multiple Choice Options

  • A n = 4
  • B n = 7
  • C n = 5
  • D n = 6

Step-by-Step Verified Solution

ΔEn=−E0Z2n2

Let it start from n to m and from m to ground.
Then 13.6×4|1−1m2|=hc30.4nm

⇒1−1m2=0.7498⇒0.25=1m2

∴ m = 2, and now 13.6×4(14−1n2)=hc108.5×10−9
n = 5

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