The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is

Question

The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is (A) 6.2 × 1015 rad s -1 (B) 4.1 × 1016 rad s -1 (C) 24 × 1010 rad s -1 (D) 9.2 × 106 rad s -1

Tardigrade Admin · Accepted Answer

Maximum angular speed will be in its ground state. Hence, ω max =(v1/r1)=(2.2 × 106/0.529 × 10-10) =4.1 × 1016 rad s -1

Q. The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is

$6.2 \times 1 0^{15} r a d s^{- 1}$

$4.1 \times 1 0^{16} r a d s^{- 1}$

$24 \times 1 0^{10} r a d s^{- 1}$

$9.2 \times 1 0^{6} r a d s^{- 1}$

Maximum angular speed will be in its ground state. Hence,
$ω_{m a x} = \frac{v _{1}}{r _{1}} = \frac{2.2 \times 1 0 ^{6}}{0.529 \times 1 0 ^{- 10}}$
$= 4.1 \times 1 0^{16} r a d s^{- 1}$

Q. The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is

6.2×1015rads−1

4.1×1016rads−1

24×1010rads−1

9.2×106rads−1

Maximum angular speed will be in its ground state. Hence, ωmax​=r1​v1​​=0.529×10−102.2×106​ =4.1×1016rads−1

$6.2 \times 1 0^{15} r a d s^{- 1}$

$4.1 \times 1 0^{16} r a d s^{- 1}$

$24 \times 1 0^{10} r a d s^{- 1}$

$9.2 \times 1 0^{6} r a d s^{- 1}$

Maximum angular speed will be in its ground state. Hence,
$ω_{m a x} = \frac{v _{1}}{r _{1}} = \frac{2.2 \times 1 0 ^{6}}{0.529 \times 1 0 ^{- 10}}$
$= 4.1 \times 1 0^{16} r a d s^{- 1}$