The number of revolutions done by an electron in one second in the first orbit of a hydrogen atom is

Question

The number of revolutions done by an electron in one second in the first orbit of a hydrogen atom is (A) 6.57× 1015 (B) 6.57× 1013 (C) 6.57× 1011 (D) 6.57× 1014

Tardigrade Admin · Accepted Answer

mathrmv1( mathrmt)= mathrmn ⋅(2 π mathrmr1) We know that frequency=(v e l o c i t y/d i s t a n c e) (v1/2 π r1)=(2.18 × 106/2 π × 0.529 × 10- 10) =6.57× 1015

Q. The number of revolutions done by an electron in one second in the first orbit of a hydrogen atom is

$6.57 \times 1 0^{15}$

$6.57 \times 1 0^{13}$

$6.57 \times 1 0^{11}$

$6.57 \times 1 0^{14}$

$v_{1} (t) = n \cdot (2 π r_{1})$
We know that $f re q u e n cy = \frac{v e l oc i t y}{d i s t an ce}$

$\frac{v _{1}}{2 π r _{1}} = \frac{2.18 \times 1 0 ^{6}}{2 π \times 0.529 \times 1 0 ^{- 10}}$
$= 6.57 \times 1 0^{15}$

Q. The number of revolutions done by an electron in one second in the first orbit of a hydrogen atom is

6.57×1015

6.57×1013

6.57×1011

6.57×1014

v1​(t)=n⋅(2πr1​) We know that frequency=distancevelocity​ 2πr1​v1​​=2π×0.529×10−102.18×106​ =6.57×1015

$6.57 \times 1 0^{15}$

$6.57 \times 1 0^{13}$

$6.57 \times 1 0^{11}$

$6.57 \times 1 0^{14}$

$v_{1} (t) = n \cdot (2 π r_{1})$
We know that $f re q u e n cy = \frac{v e l oc i t y}{d i s t an ce}$

$\frac{v _{1}}{2 π r _{1}} = \frac{2.18 \times 1 0 ^{6}}{2 π \times 0.529 \times 1 0 ^{- 10}}$
$= 6.57 \times 1 0^{15}$