For He⁺, a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm. The wavelength (in nm) of the emitted photon during the transition is:
[Use:
Bohr radius, a = 52.9 pm
Rydberg constant, R_H = 2.2 × 10⁻¹⁸ J
Planck’s constant, h = 6.6 × 10⁻³⁴ J s
Speed of light, c = 3 × 10⁸ m s⁻¹]

The correct answer is 30.

Step 1: Understand the given data
For a helium ion (He⁺):
Atomic number, Z = 2
Initial orbit radius, r_initial = 105.8 pm
Final orbit radius, r_final = 26.45 pm
Bohr radius for hydrogen atom, a₀ = 52.9 pm
Rydberg constant in terms of energy, R_H = 2.2 × 10⁻¹⁸ J
Planck's constant, h = 6.6 × 10⁻³⁴ J s
Speed of light, c = 3 × 10⁸ m s⁻¹

Step 2: Find the principal quantum numbers (n₁ and n₂)
The radius of the n-th orbit for a hydrogen-like ion is given by the formula:
$r=\frac{a_0\⁢n^2}{Z}$
Rearranging for n²:
$n^2=\frac{r\⁢Z}{a_0}$

For the initial state (r_initial = 105.8 pm):
$n_{\mathrm{initial}}^2=\frac{105.8\⁢2}{52.9}=2\⁢2=4$
$n_{\mathrm{initial}}=2$

For the final state (r_final = 26.45 pm):
$n_{\mathrm{final}}^2=\frac{26.45\⁢2}{52.9}=\frac{52.9}{52.9}=1$
$n_{\mathrm{final}}=1$

Thus, the transition occurs from n₂ = 2 to n₁ = 1.

Step 3: Calculate the energy of the emitted photon
The energy difference (ΔE) between the two energy levels for a hydrogen-like system is:
$\mathrm{\Δ}E=R_H\⁢Z^2\⁢\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Substituting n₁ = 1, n₂ = 2, Z = 2, and R_H = 2.2 × 10⁻¹⁸ J:
$\mathrm{\Δ}E=2.2\×{10}^{-18}\×2^2\×\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
$\mathrm{\Δ}E=2.2\×{10}^{-18}\×4\×\left(1-\frac{1}{4}\right)$
$\mathrm{\Δ}E=2.2\×{10}^{-18}\×4\×\frac{3}{4}$
$\mathrm{\Δ}E=6.6\×{10}^{-18}\text{\ J}$

Step 4: Calculate the wavelength (λ) of the emitted photon
Using the relation between photon energy and wavelength:
$\mathrm{\Δ}E=\frac{h\⁢c}{\mathrm{\λ}}$
$\mathrm{\λ}=\frac{h\⁢c}{\mathrm{\Δ}E}$

Substitute the given values:
$\mathrm{\λ}=\frac{\left(6.6\×{10}^{-34}\right)\×\left(3\×{10}^8\right)}{6.6\×{10}^{-18}}$
$\mathrm{\λ}=3\×{10}^{-8}\text{\ m}$

Converting the wavelength to nanometers (nm):
$\mathrm{\λ}=3\×{10}^{-8}\×{10}^9\text{\ nm}=30\text{\ nm}$

Therefore, the wavelength of the emitted photon is 30 nm.