The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 . Then the number of electrons striking the target per second is [2002]

Question

The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 $m A$ . Then the number of electrons striking the target per second is [2002]

Smart Achievers · Accepted Answer

(1)

As we know, $I = \frac{q}{t} = \frac{n e}{t}$

$∴$ No. of electrons striking the target per second

$\frac{n}{t} = \frac{I}{e} = 2 \times 10^{16}$

Solution

Q.
The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 $m A$ . Then the number of electrons striking the target per second is [2002]

1 $2 \times 10^{16}$

2 $5 \times 10^{6}$

3 $1 \times 10^{17}$

4 $4 \times 10^{15}$

Ans.
(1)

As we know, $I = \frac{q}{t} = \frac{n e}{t}$

$∴$ No. of electrons striking the target per second

$\frac{n}{t} = \frac{I}{e} = 2 \times 10^{16}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. 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]

1 2×1016

2 5×106

3 1×1017

4 4×1015

Ans. (1) As we know, I=qt=net ∴ No. of electrons striking the target per second nt=Ie=2×1016