Question

In a hydrogen tube it is observed that through a given cross-section $3.13\times {10}^{15}\text{electrons per sec}$ moving from right to left and $3.12\times {10}^{15}\text{protons per sec}$ are moving from left to right. The electric current in the discharge tube and its direction is,

• A. $1\text{mA}$ towards right
• B. $1\text{mA}$ towards left
• C. $1.6\mu \text{A}$ towards right
• D. $1.6\mu \text{A}$ towards left

Answer 1

The correct option is A $1\text{mA}$ towards right

Using the definition of current we have,

$I=\frac{n_p{q}_p}{t}+\frac{n_e{q}_e}{t}$

$=(\frac{n_p}{t}+\frac{n_e}{t})q_p$

$=\left[\right(3.12\times {10}^{15})+(3.13\times {10}^{15}\left)\right]1.6\times {10}^{-19}$

=$6.25\times {10}^{15}\times 1.6\times {10}^{-19}$

i.e. $I={10}^{16}\times {10}^{-19}={10}^{-13}\text{A}=1\text{mA}$ towards right.