In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is : [Given that Bohr radius, $a_0$ = $52.9\, pm$ ]

Question

An element 'M' has 25% of the electrons filled in the third shell as in the second shell. The element 'M' is :

Answer 1

To determine the de Broglie wavelength of an electron in the second Bohr orbit of a hydrogen atom, we will follow these steps:

The de Broglie wavelength ($\lambda$) is given by the formula: $\lambda = \frac{h}{p}$ , where:
- $h$ is Planck's constant.
- $p$ is the momentum of the electron.
For an electron in the Bohr orbit, the momentum ($p$) can also be expressed as: $p = \frac{mvr}{r} = mv$ , where:
- $m$ is the mass of the electron.
- $v$ is the velocity of the electron.
- $r$ is the radius of the orbit, specifically given as: $r = n^2a_0$.
For the second Bohr orbit ($n = 2$), the radius ($r$) is: $r_2 = 2^2a_0 = 4a_0$.
The de Broglie wavelength for the electron in the n-th orbit is: $\lambda = \frac{h}{mv} = \frac{2\pi r}{n}$.
- Substituting for $r$ in the second orbit:
- $\lambda = \frac{2\pi \times 4a_0}{2} = 4\pi a_0$
Finally, substituting the given Bohr radius $a_0 = 52.9 \, \text{pm}$:
- $\lambda = 4\pi \times 52.9 \, \text{pm} = 211.6\pi \, \text{pm}$

The correct answer is $211.6 \,\pi\, \text{pm}$, which matches the option: $211.6 \,\pi\, \text{pm}$.

The Correct Option is B