Riemann surface
matlab

Q2. Given that the ground state energy of the hydrogen atom is $-13.6 \mathrm{eV}$, the ground state energy of positronium (which is a bound state of an electron and a positron) is
(a) $+6.8 \mathrm{eV}$
(b) $-6.8 \mathrm{eV}$
(c) $-13.6 \mathrm{eV}$
(d) $-27.2 \mathrm{eV}$

Ans. : (b)
Solution: The energy expression for Positronium atom is $E_n=-\frac{13.6}{2 n^2}(\mathrm{eV})$ For $\mathrm{n}=1, E_1=\frac{-13.6}{2}(\mathrm{eV})=-6.8 \mathrm{eV}, \quad \therefore E_1=-6.8 \mathrm{eV}$

Q3. A laser operating at $500 \mathrm{~nm}$ is used to excite a molecule. If the Stokes line is observed at $770 \mathrm{~cm}^{-1}$, the approximate positions of the Stokes and the anti-Stokes lines are
(a) $481.5 \mathrm{~nm}$ and $520 \mathrm{~nm}$
(b) $481.5 \mathrm{~nm}$ and $500 \mathrm{~nm}$
(c) $500 \mathrm{~nm}$ and $520 \mathrm{~nm}$
(d) $500 \mathrm{~nm}$ and $600 \mathrm{~nm}$

Ans.:
Solution: Given $\lambda_0=500 \mathrm{~nm}=5 \times 10^{-5} \mathrm{~cm}, \bar{v}{\text {stoke }}=770 \mathrm{~cm}^{-1} \therefore \bar{v}_0=20,000 \mathrm{~cm}^{-1}$ Raman shift $\Delta \bar{v}=\bar{v}_0-\bar{v}{\text {stoke }}=19230 \mathrm{~cm}^{-1}$
Wave number of anti-stokes line is $\bar{v}{\text {anti-stake }}=\Delta \bar{v}+\bar{v}_0=39,230 \mathrm{~cm}^{-1}$ In wavelength term $\lambda{\text {anti-stoke }}=2.549 \times 10^{-7}=254.9 \mathrm{~nm}$ and $\lambda_{\text {stoke }}=12987 \mathrm{~nm}$

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