question_answer

In a neon discharge tube \[2.9\times {{10}^{18}}\,N{{e}^{+}}\] ions move to the right each second while \[1.2\times {{10}^{18}}\] electrons move to the left per second. Electron charge is \[1.6\times {{10}^{-19}}C\]. The current in the discharge tube                                                             [MP PET 1999]

A)                    1 A towards right        

B)            0.66 A towards right

C)                    0.66 A towards left    

D)            Zero

Correct Answer: B

Solution :

             Net current \[i={{i}_{+}}+{{i}_{-}}=\frac{({{n}_{+}})\,({{q}_{+}})}{t}+\frac{({{n}_{-}})\,({{q}_{-}})}{t}\] Þ \[i=\frac{({{n}_{+}})}{t}\times e+\frac{({{n}_{-}})}{t}\times e\]        \[=2.9\times {{10}^{18}}\times 1.6\times {{10}^{-19}}+1.2\times {{10}^{18}}\times 1.6\times {{10}^{-19}}\] Þ \[i=\,0.66\,A\]