The radius of the 4 text th Bohr's orbit is 0.864 nm. The de Broglie wavelength of the electron in that orbit is:

Question

The radius of the 4 text th Bohr's orbit is 0.864 nm. The de Broglie wavelength of the electron in that orbit is: (A) 13.297 mathringA (B) 1.3565 nm (C) 1.3291 pm (D) 0.1329 nm

Tardigrade Admin · Accepted Answer

2 π rn=n λ ⇒ λ=(2 π rn/n)=(2 × 3.14 × 0.864/4) nm =1.3565 nm

Q. The radius of the $4^{th}$ Bohr's orbit is $0.864 nm$ . The de Broglie wavelength of the electron in that orbit is:

$13.297 \overset{˚}{A}$

$1.3565 nm$

$1.3291 p m$

$0.1329 nm$

$2 π r_{n} = nλ$
$\Rightarrow λ = \frac{2 π r _{n}}{n} = \frac{2 \times 3.14 \times 0.864}{4} nm$
$= 1.3565 nm$

Q. The radius of the 4th Bohr's orbit is 0.864nm. The de Broglie wavelength of the electron in that orbit is:

13.297A˚

1.3565nm

1.3291pm

0.1329nm