= Spring constant from the original graph
The vertical axis of the original figure is \b[<angular frequency>], in $10^{13}\,\mathrm{rad\,s^{-1}}$, not ordinary <frequency>. Its zone-centre optical intercept is approximately $\omega_+(0)=4.8\times10^{13}\,\mathrm{s^{-1}}$. Convert the relative atomic masses using $m_u=1.66054\times10^{-27}\,\mathrm{kg}$:
$$
m_A=23m_u=3.819\times10^{-26}\,\mathrm{kg},\qquad
m_B=35.5m_u=5.895\times10^{-26}\,\mathrm{kg}.
$$
The <long-wavelength inference of diatomic-chain parameters> then gives
$$
\boxed{\alpha=\frac{m_Am_B\omega_+^2(0)}{2(m_A+m_B)}\simeq26.7\,\mathrm{N\,m^{-1}}\simeq27\,\mathrm{N\,m^{-1}}}.
$$
No factor $2\pi$ is needed because the graph already plots $\omega$.
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