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

Energy Difference in a Single-Electron System 

The radius of the \( n \)-th orbit in a single-electron system is given by:

\[ r = 52.9 \times \frac{n^2}{z} \, \text{pm} \] where \( Z \) is the atomic number.

For \( \text{He}^+ \), \( Z \) is 2. Given:

The initial radius \( r_2 = 105.8 \, \text{pm} \) and the final radius \( r_1 = 26.45 \, \text{pm} \).

For \( r_2 = 105.8 \, \text{pm} \), we have:

\[ 105.8 = 52.9 \times \frac{n_2^2}{2} \quad \Rightarrow \quad n_2 = 2 \]

For \( r_1 = 26.45 \, \text{pm} \), we have:

\[ 26.45 = 52.9 \times \frac{n_1^2}{2} \quad \Rightarrow \quad n_1 = 1 \]

The energy difference is given by:

\[ E = \frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( h \) is Planck's constant, \( c \) is the speed of light, \( R_H \) is the Rydberg constant, and \( \lambda \) is the photon's wavelength.

Substituting the known values into the equation:

\[ \frac{6.6 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3 \times 10^8 \, \text{m/s}}{\lambda} = 2.2 \times 10^{-18} \, \text{J} \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Solving for \( \lambda \), we get:

\[ \lambda = 30 \times 10^{-9} \, \text{m} \quad \text{or} \quad 30 \, \text{nm} \]

Final Answer:

The wavelength \( \lambda \) is \( \boxed{30 \, \text{nm}} \).

Answer 2

Wavelength Calculation for Photon Emission 

The radius of the orbit is given by: \[ r = 52.9 \times \frac{n^2}{z} \, \text{pm} \]

For \( r = 105.8 \, \text{pm} \), we have: \[ 105.8 = \frac{52.9 \times n^2}{2} \quad \Rightarrow \quad n_2 = 2 \]

For \( r = 26.45 \, \text{pm} \), we have: \[ 26.45 = 52.9 \times \frac{n^2}{2} \quad \Rightarrow \quad n_1 = 1 \]

Using the equation for the energy difference: \[ \Delta E = R_H h C \times z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \]

The equation for photon energy is: \[ \frac{hc}{\lambda} = R_H h C \times z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \]

Substituting the 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 \( \lambda \), we get: \[ \lambda = 300 \, \text{Å} \]

Finally, converting to nanometers: \[ \lambda = 30 \, \text{nm} \]

Final Answer:

The wavelength of the emitted photon is \( \boxed{30 \, \text{nm}} \).