Question

How many electrons must be removed from a piece of metal to give it a positive charge of 1.0×10-7C

Answer 1

$\mathtt { \huge{ \fbox{Solution :)}}}$

Given ,

Charge (q) = 1.0 × (10)^-7 C

We know that ,

The charge on a body is integral multiple of electrons

$\large \mathtt{{ \fbox{Charge \: (q) = ne}}}$

Thus ,

$\sf \hookrightarrow 1.0 \times {(10)}^{ - 7} = n \times 1.9 \times {(10)}^{ - 19} \\ \\ \sf \hookrightarrow n = \frac{1.0 \times {(10)}^{ - 7} }{1.9 \times {(10)}^{ - 19} } \\ \\ \sf \hookrightarrow n = \frac{1 \times {(10)}^{ (- 7 + 19 + 1 - 1)} }{19} \\ \\ \sf \hookrightarrow n = 0.05 \times {(10)}^{12} \\ \\ \sf \hookrightarrow n = 5 \times {(10)}^{(12 - 2)} \\ \\ \sf \hookrightarrow n = 5 \times {(10)}^{10}$

Hence , the number of electrons is 5 × (10)^10

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Answer 2

Answer:

Given ,

Charge (q) = 1.0 × (10)^-7 C

We know that ,

The charge on a body is integral multiple of electrons

\large \mathtt{{ \fbox{Charge \: (q) = ne}}}

Charge (q) = ne

Thus ,

\begin{gathered}\sf \hookrightarrow 1.0 \times {(10)}^{ - 7} = n \times 1.9 \times {(10)}^{ - 19} \\ \\ \sf \hookrightarrow n = \frac{1.0 \times {(10)}^{ - 7} }{1.9 \times {(10)}^{ - 19} } \\ \\ \sf \hookrightarrow n = \frac{1 \times {(10)}^{ (- 7 + 19 + 1 - 1)} }{19} \\ \\ \sf \hookrightarrow n = 0.05 \times {(10)}^{12} \\ \\ \sf \hookrightarrow n = 5 \times {(10)}^{(12 - 2)} \\ \\ \sf \hookrightarrow n = 5 \times {(10)}^{10}\end{gathered}

↪1.0×(10)

−7

=n×1.9×(10)

−19

↪n=

1.9×(10)

−19

1.0×(10)

−7

↪n=

19

1×(10)

(−7+19+1−1)

↪n=0.05×(10)

12

↪n=5×(10)

(12−2)

↪n=5×(10)

10

Hence , the number of electrons is 5 × (10)^10