Question

in hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is (given that Bohr radius,52.9pm)​

Answer 1

Answer:the

Explanation:

Answer 2

In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is ..

According to Bohr's theory of atomic structure, " Angular momentum of an electron in an orbit is integral multiple of  $\frac{h}{2\pi}$ ."

in mathematical form, $\bf{mvr_n=\frac{nh}{2\pi}}$

we know, linear momentum, P = mv

∴ $\bf{P=\frac{nh}{2\pi r_n}}$ ...(1)

According to De-Broglie's wavelength equation, " the wavelength of any particle is the ratio of Plank's constant to the momentum of that particle.

in mathematical form, $\bf{\lambda=\frac{h}{P}}$ ...(2)

from equations (1) and (2) we get,

$\bf{\lambda=\frac{2\pi r_n}{n}$

now putting values of n and r₂

we know, $\bf{r_n=52.9n^2}$ pm , for hydrogen atom.

so, r₂ = 52.9 × 4 = 211.6 pm

n = 2

so wavelength = $\frac{2\pi (211.6pm)}{2}$ = 211.6π pm

Therefore De-Broglie's wavelength is 211.6π pm.