Typical phonon energies are in the meV range. Typical Cooper pair energies are also in that energy ball park (or even lower). So indeed a phonon with ~meV energy can break a Cooper pair apart. However, to have sufficiently many phonons excited with enough energy to break so many Cooper pairs apart the superconducting phase is destroyed, the temperature must be high enough. At room temperature, $k_{\rm B}T \approx 25$meV, which is a lot here. At the boiling point of He, e.g., $k_{\rm B}\cdot4.2{\rm K}\approx 0.36$meV and there won't be many phonons with sufficient energies to break up a Cooper pair with say $1$meV binding energy.

Breaking Cooper pairs apart is necessary to destroy the superconducting phase; as long as Cooper pairs are bound, they are Bosonic particles obeying Bose-Einstein statistics with macroscopic ground state occupation. That ground state will only be destroyed if a significant amount of Cooper pairs is broken up.

Typical phonon energies are in the meV range. Typical Cooper pair binding energies are also in that energy ball park (or even lower). So indeed a phonon with ~meV energy can break a Cooper pair apart. However, to have sufficiently many phonons excited with enough energy to break so many Cooper pairs apart the superconducting phase is destroyed, the temperature must be high enough. At room temperature, $k_{\rm B}T \approx 25$meV, which is a lot here. At the boiling point of He, e.g., $k_{\rm B}\cdot4.2{\rm K}\approx 0.36$meV and there won't be many phonons with sufficient energies to break up a Cooper pair with say $1$meV binding energy.

Breaking Cooper pairs apart is necessary to destroy the superconducting phase; as long as Cooper pairs are bound, they are Bosonic particles obeying Bose-Einstein statistics with macroscopic ground state occupation. That ground state will only be destroyed if a significant amount of Cooper pairs is broken up.

Typical phonon energies are in the meV range. Typical Cooper pair energies are also in that energy ball park (or even lower). So indeed a phonon with ~meV energy can break a cooper pair apart. However, to have sufficiently many phonons excited with enough energy to break so many Cooper pairs apart the superconducting phase is destroyed, the temperature must be high enough. At room temperature, $k_{\rm B}T \approx 25$meV, which is a lot here. At the boiling point of He, e.g., $k_{\rm B}\cdot4.2{\rm K}\approx 0.36$meV and there won't be many phonons with sufficient energies to break up a Cooper pair with say $1$meV binding energy.

Breaking Cooper pairs apart is necessary to destroy the superconducting phase; as long as Cooper pairs are bound, they are Bosonic particles obeying Bose-Einstein statistics with macroscopic ground state occupation. That ground state will only be destroyed if a significant amount of Cooper pairs is broken up.

Typical phonon energies are in the meV range. Typical Cooper pair energies are also in that energy ball park (or even lower). So indeed a phonon with ~meV energy can break a Cooper pair apart. However, to have sufficiently many phonons excited with enough energy to break so many Cooper pairs apart the superconducting phase is destroyed, the temperature must be high enough. At room temperature, $k_{\rm B}T \approx 25$meV, which is a lot here. At the boiling point of He, e.g., $k_{\rm B}\cdot4.2{\rm K}\approx 0.36$meV and there won't be many phonons with sufficient energies to break up a Cooper pair with say $1$meV binding energy.

Breaking Cooper pairs apart is necessary to destroy the superconducting phase; as long as Cooper pairs are bound, they are Bosonic particles obeying Bose-Einstein statistics with macroscopic ground state occupation. That ground state will only be destroyed if a significant amount of Cooper pairs is broken up.