Modern Physics 4 Question 15
15. 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
$(2002,2 M)$
(a) $2 \times 10^{16}$
(b) $5 \times 10^{6}$
(c) $1 \times 10^{17}$
(d) $4 \times 10^{15}$
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Solution:
$$ i=\frac{q}{t}=\frac{n e}{t} \quad \therefore \quad n=\frac{i t}{e} $$
Substituting $i=3.2 \times 10^{-3} A$,
$$ \begin{aligned} e & =1.6 \times 10^{-19} C \text { and } t=1 s \\ \text { we get, } \quad n & =2 \times 10^{16} \end{aligned} $$
