The de-Broglie’s wavelength of electron present in first Bohr orbit of ‘H’ atom is :

Question

The de-Broglie’s wavelength of electron present in first Bohr orbit of ‘H’ atom is : (A) 0.529 mathringA (B) 2 π × 0.529 mathringA (C) (0.529/2π) mathringA (D) 4 × 0.529 mathringA

Tardigrade Admin · Accepted Answer

Bohr's radius of first stationary state is r=0.529 mathringA Angular momentum is given by mvr=(n h/2 π) For first Bohr's orbit n=1 2 π r =(h/m v)=λ λ =2 π × 0.529 mathringA

Q. The de-Broglie’s wavelength of electron present in first Bohr orbit of ‘H’ atom is :

$0.529 \overset{˚}{A}$

$2 π \times 0.529 \overset{˚}{A}$

$\frac{0.529}{2 π} \overset{˚}{A}$

$4 \times 0.529 \overset{˚}{A}$

Bohr's radius of first stationary state is $r = 0.529 \overset{˚}{A}$

Angular momentum is given by $m v r = \frac{nh}{2 π}$

For first Bohr's orbit $n = 1$

$2 π r = \frac{h}{m v} = λ$

$λ = 2 π \times 0.529 \overset{˚}{A}$

Q. The de-Broglie’s wavelength of electron present in first Bohr orbit of ‘H’ atom is :

0.529A˚

2π×0.529A˚

2π0.529​A˚

4×0.529A˚

Bohr's radius of first stationary state is r=0.529A˚ Angular momentum is given by mvr=2πnh​ For first Bohr's orbit n=1 2πr=mvh​=λ λ=2π×0.529A˚

$0.529 \overset{˚}{A}$

$2 π \times 0.529 \overset{˚}{A}$

$\frac{0.529}{2 π} \overset{˚}{A}$

$4 \times 0.529 \overset{˚}{A}$

Bohr's radius of first stationary state is $r = 0.529 \overset{˚}{A}$

Angular momentum is given by $m v r = \frac{nh}{2 π}$

For first Bohr's orbit $n = 1$

$2 π r = \frac{h}{m v} = λ$

$λ = 2 π \times 0.529 \overset{˚}{A}$