Q.The common-base DC current gain of a transistor is 0.967. If the emitter current is 10mA. What is the value of base current?

For a transistor in the common emitter configuration, the relationship between its three currents ($I_E=I_B+I_C$) leads to two related current-gain ratios: the common-emitter DC current amplification factor $\beta_{dc}=\dfrac{I_C}{I_B}$ (typically 20 to 200 for general-purpose transistors) and the common-base DC current gain $\alpha_{dc}=\dfrac{I_C}{I_E}$ (always a fraction just under 1). Dividing $I_E=I_B+I_C$ through by $I_E$ and rearranging gives the standard relations linking them, $\beta_{dc}=\dfrac{\alpha_{dc}}{1-\alpha_{dc}}$ and equivalently $\alpha_{dc}=\dfrac{\beta_{dc}}{1+\beta_{dc}}$ -- letting either ratio be found directly from the other for a given transistor.

A transistor's CE behaviour is captured experimentally by two standard characteristic curves. The INPUT characteristic plots base current $I_B$ against base-emitter voltage $V_{BE}$ at constant collector-emitter voltage $V_{CE}$, resembling an ordinary forward-diode curve (since the base-emitter junction is a forward-biased junction); from it, the input dynamic resistance is $r_i=(\Delta V_{BE}/\Delta I_B)_{V_{CE}\,\text{const}}$. The OUTPUT characteristic plots collector current $I_C$ against collector-emitter voltage $V_{CE}$ at constant base current $I_B$, rising steeply near the origin before flattening into a nearly horizontal line -- $I_C$ becomes essentially independent of $V_{CE}$ once the collector-base junction is properly reverse biased, and instead scales with $I_B$; from it, the output dynamic resistance is $r_o=(\Delta V_{CE}/\Delta I_C)_{I_B\,\text{const}}$, which comes out comparatively LARGE because the curve is nearly flat. Together, a small $r_i$ and a large $r_o$ are exactly what gives the common emitter configuration its large overall power gain.