Question

The spin-orbit effect splits the \( ^2p \rightarrow ^2s \) transition (wavelength, \( \lambda = 6521 \, \text{Å} \)) in Lithium into two lines with separation of \( \Delta \lambda = 0.14 \, \text{Å} \). The corresponding positive value of energy difference between the above two lines, in eV, is \( m \times 10^{-5} \). The value of \( m \) (rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\). (Given: Planck's constant, \( h = 4.125 \times 10^{-15} \, \text{eV s} \), speed of light, \( c = 3 \times 10^8 \, \text{m/s} \))

Hint:

To calculate the energy difference between two spectral lines due to spin-orbit effects, use the formula \( \Delta E = \frac{h c \Delta \lambda}{\lambda^2} \).

Answer 1

Correct Answer: 3

The energy difference between the two lines can be calculated using the formula: \[ \Delta E = \frac{h c \Delta \lambda}{\lambda^2}. \] Substitute the given values: \[ \Delta E = \frac{4.125 \times 10^{-15} \times 3 \times 10^8 \times 0.14 \times 10^{-10}}{(6521 \times 10^{-10})^2}. \] This gives: \[ \Delta E \approx 3 \times 10^{-5} \, \text{eV}. \] Thus, the value of \( m \) is 3.