A cathode emits 1.8×1017 electrons/s and all the electrons reach the anode when it is given a positive potential of 400 V. Given e=1.6×10−19C, the maximum anode current is
28.8 mA
6.4 mA
2.88 mA
7.2 mA
A
2.88 mA
B
28.8 mA
C
6.4 mA
D
7.2 mA
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Solution
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The correct option is B 28.8 mA maximum anode corrent = n x e = 1.8∗1017∗1.6∗10−19 = 2.88∗10−2A = 28.8∗10−3A = 28.8mA.
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