GATE 2023 PH – Question 47
For ions of mass m, $U(r)=-\epsilon(r/r_0)e^{-r/r_0}$. What is the Einstein frequency in Hz?
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Correct answer: (B) $\sqrt{\epsilon/(mer_0^2)}/(2\pi)$
Explanation
**Einstein model.** Each ion vibrates about its equilibrium separation as an independent harmonic oscillator, with the spring constant given by the curvature of the potential at the minimum.
**Position of the minimum.** With $U(r)=-\epsilon\dfrac{r}{r_0}e^{-r/r_0}$ and $x=r/r_0$:
$$\frac{dU}{dr}=-\frac{\epsilon}{r_0}e^{-x}(1-x)=0\;\Rightarrow\;x=1,\ \text{i.e. }r=r_0 .$$
**Curvature at the minimum:**
$$\frac{d^2U}{dr^2}=-\frac{\epsilon}{r_0^2}\left[-e^{-x}(1-x)-e^{-x}\right]=\frac{\epsilon}{r_0^2}e^{-x}(2-x),$$
and at $x=1$ this is $\dfrac{\epsilon}{e\,r_0^2}$.
**Frequency:**
$$f=\frac1{2\pi}\sqrt{\frac{U^{\prime\prime}}{m}}=\frac1{2\pi}\sqrt{\frac{\epsilon}{m\,e\,r_0^2}}\quad(\text{option B}).$$