GATE 2021 PH – Question 65
Spin–orbit interaction splits the $^2P\to{}^2S$ transition of lithium, of wavelength 6521 Å, into two lines separated by $\Delta\lambda=0.14$ Å. Their positive energy separation is $m\times10^{-5}$ eV. Find $m$, rounded to the nearest integer. Given $h=4.125\times10^{-15}$ eV s and $c=3\times10^8$ m s⁻¹.
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Correct answer: 3 to 5
Explanation
Photon energy is $E=hc/\lambda$. For the small wavelength separation, $|\Delta E|\simeq hc|\Delta\lambda|/\lambda^2$. Convert $\lambda=6.521\times10^{-7}$ m and $\Delta\lambda=1.4\times10^{-11}$ m. Then $|\Delta E|=(4.125\times10^{-15})(3\times10^8)(1.4\times10^{-11})/(6.521\times10^{-7})^2=4.074\times10^{-5}$ eV. Thus $m\simeq4.07$, which rounds to 4; the official accepted interval is 3–5.