Question

4.
The radius of the 4th Bohr's orbit is 0.864 nm. The
de Broglie wavelength of the electron in that orbit
is :
(a) 13.297 Å
(b) 1.3565 nm
(c) 1.3291 pm (d) 0.1329nm​

Answer 1

Answer:

d is correct

Explanation:

good morning

Answer 2

The de Broglie wavelength of the electron in that orbit is (b) 1.3565nm.

Given:

The radius of the 4th Bohr's orbit is 0.864 nm.

To Find:

The de Broglie wavelength of the electron in that orbit.

Solution:

To find the de Broglie wavelength of the electron in that orbit we will follow the following steps:

As we know,

According to the postulate of Bohr's,

2πR = nd

$d = \frac{2\pi \: r}{n}$

Here, n is Bohr's orbit which is given as 4.

r is radius = 0.864 nm.

Now, putting values in the above formula we get,

$d = \frac{2 \times 3.14 \times \: 0.864}{4} = 1.3565nm$

Here, π is taken = 3.14

Henceforth, the de Broglie wavelength of the electron in that orbit is (b) 1.3565 nm.