The de Broglie wavelength of an electron in 2nd orbit is ( a0 = Bohr radius )

Question

The de Broglie wavelength of an electron in 2nd orbit is ( a0 = Bohr radius ) (A) 2π a0 (B) 4π a0 (C) 8π a0 (D) 16π a0

Tardigrade Admin · Accepted Answer

Radius of nth orbit is rn = n2 a0 where a0 is, the Bohr radius. If λ is the de Broglie wavelength of an electron while revolving in nth orbit of radius rn, then 2π rn = nλ; λ = (2π rn/n) For 2nd orbit, n = 2 ∴ λ = (2π rn/n) = (2π(2)2 a0/2) = 4π a0

Q. The de Broglie wavelength of an electron in $2^{n d}$ orbit is ( $a_{0}$ = Bohr radius )

$2 π a_{0}$

$4 π a_{0}$

$8 π a_{0}$

$16 π a_{0}$

Q. The de Broglie wavelength of an electron in 2nd orbit is ( a0​ = Bohr radius )

2πa0​

4πa0​

8πa0​

16πa0​

Radius of nth orbit is rn​=n2a0​ where a0​ is, the Bohr radius. If λ is the de Broglie wavelength of an electron while revolving in nth orbit of radius rn​, then 2πrn​=nλ;λ=n2πrn​​ For 2nd orbit, n=2 ∴λ=n2πrn​​=22π(2)2a0​​=4πa0​

Q. The de Broglie wavelength of an electron in $2^{n d}$ orbit is ( $a_{0}$ = Bohr radius )

$2 π a_{0}$

$4 π a_{0}$

$8 π a_{0}$

$16 π a_{0}$