Two electrons are moving towards each other, each with a velocity of 106 m / s. What will be closest distance of approach between them?

Question

Two electrons are moving towards each other, each with a velocity of 106 m / s. What will be closest distance of approach between them? (A) 1.53 × 10-8 m (B) 2.53 × 10-10 m (C) 2.53 × 10-6 m (D) Zero

Tardigrade Admin · Accepted Answer

(1/2)(9.1 × 10-31)(106)2 =(9 × 109 ×(1.6 × 10-19)2/r) 9.1 × 10-19=(9 × 109 × 2.56 × 10-38/r) r=2.56 × 10-10 m

Q. Two electrons are moving towards each other, each with a velocity of $1 0^{6} m / s$ . What will be closest distance of approach between them?

$1.53 \times 1 0^{- 8} m$

$2.53 \times 1 0^{- 10} m$

$2.53 \times 1 0^{- 6} m$

Zero

$\frac{1}{2} (9.1 \times 1 0^{- 31}) (1 0^{6})^{2}$
$= \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{r}$
$9.1 \times 1 0^{- 19} = \frac{9 \times 1 0 ^{9} \times 2.56 \times 1 0 ^{- 38}}{r}$
$r = 2.56 \times 1 0^{- 10} m$

Q. Two electrons are moving towards each other, each with a velocity of 106m/s. What will be closest distance of approach between them?

1.53×10−8m

2.53×10−10m

2.53×10−6m