Question

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, 𝑅H = 2.2 × 10⁻¹⁸ J, Planck’s constant, h = 6.6 × 10⁻³⁴ J s, Speed of light, c = 3 × 10⁸ m s⁻¹]

Answer 1

Correct Answer: 30

Photon Wavelength Calculation 

Given: The formula for the radius \( r \) is:

\(r = 52.9 \times \frac{n^2}{z} \,\, \text{pm}\)

From the equation:

\(\therefore 105.8 = \frac{52.9 \times n^2}{2} \,\,\, \therefore n_2 = 2\)

Similarly, for \( n_1 \):

\(26.45 = 52.9 \times \frac{n^2}{2} \,\,\, \therefore n_1 = 1\)

Energy Change Formula:

The energy change \( \Delta E \) is given by:

\(\Delta E = R_H h C \times z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]\)

Using the equation for wavelength, we have:

\(\frac{hc}{\lambda} = R_H h C \times z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]\)

Substituting the known values:

\(\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{\lambda} = 2.2 \times 10^{-18} \times 4 \times \frac{3}{4}\)

Solving for \( \lambda \):

\(\therefore \lambda = 300 \, \text{Å}\)

Finally, converting to nanometers:

\(\therefore \lambda = 30 \, \text{nm}\)

The wavelength of the emitted photon is 30 nm.