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11.Dual Nature of Radiation and matter
medium
A cathode emits $1.8 \times {10^{14}}$ electrons per second, when heated. When $ 400\;V$ is applied to anode all the emitted electrons reach the anode. The charge on electron is $1.6 \times {10^{ - 19}}\;C$. The maximum anode current is ............ $\mu A$
A
$2.7$
B
$29$
C
$72$
D
$2.9$
Solution
(b) $i = \frac{Q}{t} = \frac{{ne}}{t} = 1.8 \times {10^{14}} \times 1.6 \times {10^{ – 19}} = 28.8 \times {10^{ – 6}}A$
$ = 29\,\mu A$
Standard 12
Physics
