Two electrons are approaching each other with a relative velocity of 106 m s-1 . The closest distance of their approach is

Question

Two electrons are approaching each other with a relative velocity of 106 m s-1 . The closest distance of their approach is (A)  0.51 nm  (B)  0.51 mm  (C)  0.51 mathring A  (D) 0.51 μ m

Tardigrade Admin · Accepted Answer

Let r be distance of closest approach. v = 106 m s-1 ∴ Kinetic energy = Potential energy between the two electrons or (1/2) me v2 = (1/4 π e0) = (q1q2/r) r = (2 q1q2/4 π e0) ×(1/me v2) = (2 × (1.6 × 10-19)2× 9 × 109/9.1 × 10-31× (106)2) =5.1 × 10-10 =0.51 × 10-9 = 0.51 nm

Q. Two electrons are approaching each other with a relative velocity of $1 0^{6} m s^{- 1}$ . The closest distance of their approach is

$0.51 nm$

$0.51 mm$

$0.51 \overset{˚}{A}$

$0.51 μ m$

Q. Two electrons are approaching each other with a relative velocity of 106ms−1 . The closest distance of their approach is

0.51nm

0.51mm

0.51A˚

0.51μm

Let r be distance of closest approach. v=106ms−1 ∴ Kinetic energy = Potential energy between the two electrons or 21​me​v2=4πe0​1​=rq1​q2​​ r=4πe0​2q1​q2​​×me​v21​ =9.1×10−31×(106)22×(1.6×10−19)2×9×109​ =5.1×10−10=0.51×10−9=0.51nm

Q. Two electrons are approaching each other with a relative velocity of $1 0^{6} m s^{- 1}$ . The closest distance of their approach is

$0.51 nm$

$0.51 mm$

$0.51 \overset{˚}{A}$

$0.51 μ m$