Question

A cathode emits $1.8\times {10}^{17}$ electrons/s and all the electrons reach the anode when it is given a positive potential of 400 V. Given $e=1.6\times {10}^{-19}$C, the maximum anode current is

• A. 2.88 mA
• B. 28.8 mA
• C. 7.2 mA
• D. 6.4 mA

Answer 1

Answer: (B)

28.8 mA

maximum anode corrent
= n x e
= $1.8\ast {10}^{17}\ast 1.6\ast {10}^{-19}$
= $2.88\ast {10}^{-2}A$
= $28.8\ast {10}^{-3}A$
= $28.8mA$.