In a Millikan's oil drop experiment the charge on an oil drop is calculated to be 6.35 × 10-19 C. The number of excess electrons on the drop is

Question

In a Millikan's oil drop experiment the charge on an oil drop is calculated to be 6.35 × 10-19 C. The number of excess electrons on the drop is (A) 32 (B) 4 (C) 42 (D) 6

Tardigrade Admin · Accepted Answer

As we know, Q = ne Number of electron n=(Q/e) (6.35×10-19/1.6×10-19) =3.9=4

Q. In a Millikan's oil drop experiment the charge on an oil drop is calculated to be $6.35 \times 1 0^{- 19} C$ . The number of excess electrons on the drop is

$32$

$4$

$42$

$6$

As we know, $Q = n e$
Number of electron $n = \frac{Q}{e}$
$\frac{6.35 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}}$
$= 3.9 = 4$

Q. In a Millikan's oil drop experiment the charge on an oil drop is calculated to be 6.35×10−19C. The number of excess electrons on the drop is

32

4

42

6

As we know, Q=ne Number of electron n=eQ​ 1.6×10−196.35×10−19​ =3.9=4

Q. In a Millikan's oil drop experiment the charge on an oil drop is calculated to be $6.35 \times 1 0^{- 19} C$ . The number of excess electrons on the drop is

$32$

$4$

$42$

$6$

As we know, $Q = n e$
Number of electron $n = \frac{Q}{e}$
$\frac{6.35 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}}$
$= 3.9 = 4$