An excited He$$^+$$ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm in making a transition to the ground state. The quantum number n, corresponding to its initial excited state is
(for a photon of wavelength $$\lambda$$, energy E = $$\frac{1240 \text{ eV}}{\lambda(\text{in nm})}$$)

Given:
Energy of photon,

$$E=\ \frac{1240​eV\ }{λ(innm)}$$

Find the initial principal quantum number (n) of the ion.

Solution

For a hydrogen-like ion (He⁺, Z=2):

$$\ \frac{\ 1}{λ}​=R\cdot Z^2(\ \frac{\ 1}{n_f^2}​−\ \frac{\ 1}{n_i^2}​)$$

Two successive transitions:

1. $$From\ n→m\ with\ λ_1​=108.5nm$$

$$\ \frac{\ 1}{108.5}​=R\cdot(2)^2(\ \frac{\ 1}{m^2}​−\ \frac{\ 1}{n^2}​)$$

  2. $$From\ m→1\ withλ_2​=30.4nm$$

$$\ \frac{\ 1}{30.4}​=R(2)^2(1−\ \frac{\ 1}{m^2}​)$$

Solving, we get:

m=2

Substitute into first equation:

$$\ \frac{\ 1}{108.5}=4R(\ \frac{\ 1}{4}−\ \frac{\ 1}{n^2})$$

$$⇒n=5$$