The drift velocity of electrons in a metal is given by the equation

$I=enAv_D$

where $I$ is the electric current in the metal wire, $n$ is the number of electron density, $A$ is the cross sectional area of the metal wire and $v_D$ is the drift velocity.

From this we get

$v_D= \frac{I}{enA}$

The thermal velocity is given by

$\frac{1}{2}m_ev_T^2=\frac{3}{2}k_BT$

where $T$ is the temperature of the metal, $k_B$ is Boltzmann’s constant, $m_e$ is electron mass and $v_T$ is the thermal velocity. From the last equation we get

$v_T= \sqrt{\frac{ek_BT}{m_e}}$

Equating $v_D$ and $v_T$ gives the following

condition for the two speeds top be equal:

$\frac{I}{enA }=\sqrt{\frac{ek_BT}{m_e}}$.

The drift velocity of electrons in a metal is given by the equation

$I=enAv_D$

where $I$ is the electric current in the metal wire, $n$ is the number of electron density, $A$ is the cross sectional area of the metal wire and $v_D$ is the drift velocity.

From this we get

$v_D= \frac{I}{enA}$

The thermal velocity is given by

$\frac{1}{2}m_ev_T^2=\frac{3}{2}k_BT$

where $T$ is the temperature of the metal, $k_B$ is Boltzmann’s constant, $m_e$ is electron mass and $v_T$ is the thermal velocity. From the last equation we get

$v_T= \sqrt{\frac{3k_BT}{m_e}}$

Equating $v_D$ and $v_T$ gives the following

Condition for the two speeds to be equal:

$\frac{I}{enA }=\sqrt{\frac{3k_BT}{m_e}}$.