calculate the no. of electrons present in 1 colume in charge . please write the steps also.
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Answered by
2
★Ello ★
here is your answer!!
Q = ne
q = quantization
n = number of electons
e = mass of electons
1 = 1.6 × 10 ^- 19 × number of electons
hence the number of electons =>
1 / 1.6 × 10 ^ -19
or 6 .25 × 10 ^ 18 electrons
hence your answer is
★ 6 .25 × 10 ^ 18 electrons ★
hope it helps you dear friend !!!
here is your answer!!
Q = ne
q = quantization
n = number of electons
e = mass of electons
1 = 1.6 × 10 ^- 19 × number of electons
hence the number of electons =>
1 / 1.6 × 10 ^ -19
or 6 .25 × 10 ^ 18 electrons
hence your answer is
★ 6 .25 × 10 ^ 18 electrons ★
hope it helps you dear friend !!!
yashbarik:
you have to take -19
hey may i
the answer to the question is 6.022 *10to the power of the 20
Answered by
13
hey dear here is your answer
given : q= 1 c , e = 1.6 × 10^-19 c
let the number of electrons constituting 1 c of charge be n..
using q= ne , we get
n = q/ e
n= 1/1.6×10^-19
= n= 6.25×10^18
hence , no of electrons is 6.25×10^18 e
given : q= 1 c , e = 1.6 × 10^-19 c
let the number of electrons constituting 1 c of charge be n..
using q= ne , we get
n = q/ e
n= 1/1.6×10^-19
= n= 6.25×10^18
hence , no of electrons is 6.25×10^18 e
can you explain 2nd last step
yes
here n = 1 / 1.6× 10^-19 and when 10^-19 goes up it becomes 1×10^19 / 1.6 then you can take value that 10 ×10^18/1.6 on dividing 10 by 1.6 you will get 6.25 the answer is 6.25×10^18
thanks
my pleasure
the answer is 6.022*10to the power of the 20
hey dear it is not answer
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