An electron is revolving around a nucleus at a distance 0.528 mathringA with velocity 2.18 × 10-6m s-1 .The acceleration of electron is

Question

An electron is revolving around a nucleus at a distance 0.528 mathringA with velocity 2.18 × 10-6m s-1 .The acceleration of electron is (A) zero (B) 9×1022m s-2 (C) 1020 m s-2 (D) 7.6×1021m s-2

Tardigrade Admin · Accepted Answer

Using centripetal acceleration, a = ( upsilon2/r) we get, a = ((2.18 × 106) (2.18 × 106)/0.528× 10-10) = 9 × 1022 m s-1

Q. An electron is revolving around a nucleus at a distance $0.528 \overset{˚}{A}$ with velocity $2.18 \times 1 0^{- 6} m s^{- 1}$ .The acceleration of electron is

zero

$9 \times 1 0^{22} m s^{- 2}$

$1 0^{20} m s^{- 2}$

$7.6 \times 1 0^{21} m s^{- 2}$

Using centripetal acceleration,
$a = \frac{υ ^{2}}{r}$ we get,
$a = \frac{( 2.18 \times 1 0 ^{6} ) ( 2.18 \times 1 0 ^{6} )}{0.528 \times 1 0 ^{- 10}} = 9 \times 1 0^{22} m s^{- 1}$

Q. An electron is revolving around a nucleus at a distance 0.528A˚ with velocity 2.18×10−6ms−1 .The acceleration of electron is

zero

9×1022ms−2

1020ms−2

7.6×1021ms−2

Using centripetal acceleration, a=rυ2​ we get, a=0.528×10−10(2.18×106)(2.18×106)​=9×1022ms−1

Q. An electron is revolving around a nucleus at a distance $0.528 \overset{˚}{A}$ with velocity $2.18 \times 1 0^{- 6} m s^{- 1}$ .The acceleration of electron is

$9 \times 1 0^{22} m s^{- 2}$

$1 0^{20} m s^{- 2}$

$7.6 \times 1 0^{21} m s^{- 2}$

Using centripetal acceleration,
$a = \frac{υ ^{2}}{r}$ we get,
$a = \frac{( 2.18 \times 1 0 ^{6} ) ( 2.18 \times 1 0 ^{6} )}{0.528 \times 1 0 ^{- 10}} = 9 \times 1 0^{22} m s^{- 1}$