= Solution
An <electron> losing <energy> to an acoustic <phonon> has, in a normal process, $\mathbf k'=\mathbf k_F-\mathbf k_{\rm ph}$, with possible addition of a reciprocal-lattice vector for an <Umklapp process>. The <energy> transfer is at most of order $\hbar\omega_D=k_B\theta_D$, much smaller than $E_F=k_BT_F$. Hence the magnitudes change very little even though the <momentum> direction can change substantially. At $T\sim\theta_D\ll T_F$, thermal vacancies near the <Fermi surface> allow emission into some lower-energy states; exactly at zero <temperature> emission into an occupied Fermi sea would be blocked by the <Pauli exclusion principle>.
Using $k'=(1-\delta)k_F$ and the free-electron dispersion, <energy> conservation gives
$$
\frac{\hbar^2k_F^2}{2m_e}\left[1-(1-\delta)^2\right]
=\hbar c k_{\rm ph}.
$$
Since $\delta\ll1$, the bracket is $2\delta+O(\delta^2)$. Rearranging yields
$$
\boxed{\frac1{2\delta}\frac{k_{\rm ph}}{k_F}\frac1{k_F}
\simeq\frac\hbar{2m_ec}=\lambda.}
$$
Equivalently $\delta\simeq\hbar c k_{\rm ph}/(2E_F)$. For a Debye-cutoff <phonon> in the preceding model, $\delta\simeq\theta_D/(2T_F)=1/480=0.00208$, confirming the approximation. Near-elastic scattering constrains the radial distance in wave-vector space, not the <scattering angle> alone.
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