In a bipolar transistor at room temperature if the emitter current is doubled the voltage its junction across base-emitter (for η = 1): 

In a BJT emitter current is doubled and one has to find the base emitter voltage, so we consider the p-n junction formed by emitter and base. The junction current is given by

\(\rm I_e=I_0(e^{Vb_e/η V_T}-1)\)

Given the current is doubled 

So, \(\rm \frac{2I_1}{I_1}=\frac{I_0[e^{Vb_{e2}/η V_T}-1]}{I_0[e^{Vb_{e1}/η V_T}-1]}\)

or \(\rm \frac{2}{1}=\frac{e^{Vb_{e2}/η V_T}-1}{e^{Vb_{e2}/η V_T}-1}\)

Since, \(\rm e^{Vb_{e2}/η V_T}>>1\)

\(\rm e^{Vb_{e1}/η V_T}>>1\)

Then \(\rm e^{(V_{be2}-V_{be1})/η V_T}=2\)

or, \(\rm (V_{be2}-V_{be1})=η V_T ln2\)

Taking η =1

V_be2 - V_be1 = 1 × 0.026 × 0.693

= 18 mV