Millikan's oil drop experiment attempts to measure the charge on a single electron, e, by measuring the charge of tiny oil drops suspended in an electrostatic field. It is assumed that the charge on the oil drop is due to just a small number of excess electrons. The charges 3.90 × 10-19 C , 6.50 × 10-19 C and 9.10 × 10-19 C are measured on three drops of oil. The charge of an electron is deduced to be,

Question

Millikan's oil drop experiment attempts to measure the charge on a single electron, e, by measuring the charge of tiny oil drops suspended in an electrostatic field. It is assumed that the charge on the oil drop is due to just a small number of excess electrons. The charges 3.90 × 10-19 C , 6.50 × 10-19 C and 9.10 × 10-19 C are measured on three drops of oil. The charge of an electron is deduced to be, (A) 1.3 × 10-19 C (B) 1.6 × 10-19 C (C) 2.6 × 10-19 C (D) 3.9 × 10-19 C

Tardigrade Admin · Accepted Answer

3.9 × 10-19= n 1 e , 6.5 × 10-19= n 2 e , 9.1 × 10-19= n 3 e n 1, n 2 and n 3 are integers for e =1.3 × 10-19

$1.3 \times 1 0^{- 19} C$

$1.6 \times 1 0^{- 19} C$

$2.6 \times 1 0^{- 19} C$

$3.9 \times 1 0^{- 19} C$

$3.9 \times 1 0^{- 19} = n_{1} e,$
$6.5 \times 1 0^{- 19} = n_{2} e,$
$9.1 \times 1 0^{- 19} = n_{3} e$
$n_{1}, n_{2}$ and $n_{3}$
are integers for $e = 1.3 \times 1 0^{- 19}$

1.3×10−19C

1.6×10−19C

2.6×10−19C

3.9×10−19C

3.9×10−19=n1​e, 6.5×10−19=n2​e, 9.1×10−19=n3​e n1​,n2​ and n3​ are integers for e=1.3×10−19

$1.3 \times 1 0^{- 19} C$

$1.6 \times 1 0^{- 19} C$

$2.6 \times 1 0^{- 19} C$

$3.9 \times 1 0^{- 19} C$

$3.9 \times 1 0^{- 19} = n_{1} e,$
$6.5 \times 1 0^{- 19} = n_{2} e,$
$9.1 \times 1 0^{- 19} = n_{3} e$
$n_{1}, n_{2}$ and $n_{3}$
are integers for $e = 1.3 \times 1 0^{- 19}$