Question

Please solve it urgently ​

Answer 1

ANSWER

${6.25 \times {10}^{11} }$

Step By Step Explanation

We have,

Total charge present on metal, Q= 1× $10^-7$c

A number of total electrons to be removed=?

Let, it be n.

And charge on one electron is,e = 1.6× ${10}^{-19}$c

Now we know that=>

Q = n × e. --------------(i)

and Q = I × t. ------------------(ii)

Now, we shall use the first formula(i) to solve this =>

Q = n × e

=>n = $\dfrac{Q}{e}$

=>n = $\dfrac{1\times {10}^{-7} }{1.6 \times {10}^ {-19} }$

=>n = $\dfrac{1\times {10}^{-7+19} }{1.6}$

{By using law of exponents}

=>n = $\dfrac{1 \times {10}^{12} }{1.6}$

=>n = ${0.625 \times {10}^{12} }$

=>n = ${6.25 \times {10}^{11} }$

So, ${6.25 \times {10}^{11} }$ number of electrons are needed to remove from the metal.

Remember

1)Q = n × e.

2)Q = I × t.

3)$a^{n} \times a^{m}=a^{m+n}$

4)$a^{m}\div a^{n}= a^{m-n}$

5)${a^m }^n= a^{mn}$

6)$a^{0} = 1$

7) $(\dfrac{a}{b})^n= \dfrac{{a}^n}{{b}^n}$

8)$a^{-1}=\dfrac{1}{a}$

9)$a^{x/y}={y\sqrt{a}}^x$