**Question: How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F= 96,500 C/mol) (CBSE 2017) **

Solution:

The value of current (I) 0.5 A and time (t) is given, so using the values in the formula

**Q= I.t** = 0.5 x (2 x 60 x 60) = ½ x 2 x 60 x 60 = 3600 C

A flow of 1 mole of electrons is 1F that is 96500 C. Also, 1 mole of electrons is 6.023 x 10^{23} electrons.

Number of electrons that flow in 3600 C is

= (3600 x 6.023 x 10^{23})/ 96500

**= 2.246 x 10 ^{22} electrons **

**Solving the numerical problem using the log table, the rules applied are -**

Log (m x n) = Log m + Log n

Log (m/n) = Log m - Log n

Log (m^{n})= n Log m

From the log table, the values are-

Log x = (Log 3600 + log 6.023 + 23 log 10) – (log 96500)

Log 3600 = 3.5563

Log 6.023 = 0.7798

Log 10 = 1

Log 96500= 4.9800

Log x = (3.5563 + 0.7798 + 23) – 4.9800

Separating the characteristic and mantissa,

= (3 + 0.5563 + 0.7798 + 23) - 4.9800

= 26 + 1.3361 - 4.9800 = 26 + 1 + 0.3361 - 4 - 0.9800 = (26 + 1 - 4) + 0.3361 - 0.9800

= 23 - 0.6439 + 1 - 1 (to get a positive number, add 1 and subtract 1)

= 23-1 + 1- 0.6439 = 22 + 0.3561 = 22.3561

x = antilog (22.3561) =** 2.271 x 10 ^{22} electrons**

**Therefore, 2.271 x 10 ^{22} electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours.**