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Physics Question on Electric Current

The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is

A

2×10162 \times 10^{16}

B

5×1065 \times 10^{6}

C

1×10171 \times 10^{17}

D

4×10154 \times 10^{15}

Answer

2×10162 \times 10^{16}

Explanation

Solution

  i=qt=net  ∴  n=ite\, \, i = \frac{q}{t} = \frac{ne}{t}\, \, \therefore\, \, n = \frac{it}{e}
Substituting i=3.2×10−3A,        e=1.6×10−19C  and  t=1si = 3.2 \times 10^{-3} A, \, \, \, \, \, \, \, \, e = 1.6 \times 10^{-19} C \, \, and \, \, t = 1s
we get, n=2×1016n = 2 \times 10^{16}