What is the velocity of electron present in first Bohr orbit of hydrogen atom?

Question

What is the velocity of electron present in first Bohr orbit of hydrogen atom? (A) 2.18 × 105 m/s (B) 2.18 × 106 m/s (C) 2.18 × 10-18 m/s (D) 2.18 × 10-9 m/s

Tardigrade Admin · Accepted Answer

From Bohrâs theory, v=2.18×106×(Z/n) m/s =2.1×106×(1/1) =2.18×106 m/s

Q. What is the velocity of electron present in first Bohr orbit of hydrogen atom?

$2.18 \times 1 0^{5} m / s$

$2.18 \times 1 0^{6} m / s$

$2.18 \times 1 0^{- 18} m / s$

$2.18 \times 1 0^{- 9} m / s$

From Bohr’s theory, $v = 2.18 \times 1 0^{6} \times \frac{Z}{n} m / s$
$= 2.1 \times 1 0^{6} \times \frac{1}{1}$
$= 2.18 \times 1 0^{6} m / s$

Q. What is the velocity of electron present in first Bohr orbit of hydrogen atom?

2.18×105m/s

2.18×106m/s

2.18×10−18m/s

2.18×10−9m/s

From Bohr’s theory, v=2.18×106×nZ​m/s =2.1×106×11​ =2.18×106m/s

$2.18 \times 1 0^{5} m / s$

$2.18 \times 1 0^{6} m / s$

$2.18 \times 1 0^{- 18} m / s$

$2.18 \times 1 0^{- 9} m / s$

From Bohr’s theory, $v = 2.18 \times 1 0^{6} \times \frac{Z}{n} m / s$
$= 2.1 \times 1 0^{6} \times \frac{1}{1}$
$= 2.18 \times 1 0^{6} m / s$