Question

If the de Broglie wavelength of the electron in the ${{n}^{th}}$ Bohr orbit in the hydrogen atom is equal to 1.5 $\pi {{a}_{0}}$( ${{a}_{0}}$ is the Bohr's radius), then the value of n/z is:
(A) 1.0
(B) 0.75
(C) 0.40
(D) 1.50

Answer 1

Hint: In de Broglie’s explanation of the second postulate of Bohr, there is an assumption created that the integral number of wavelengths must be equal to the circumference of the circular orbit. The integral multiple is found to be equal to the quantization number. This will help us to solve the question.

Complete step by step solution:
The postulate states that the integral number of the wavelengths will be fitted in the circumference of the orbit, then the quantization number will be equal to the integral multiple.
Therefore the equation can be written as:
\[2\pi {{r}_{n}}=n\lambda \]
Given in the question:
De Broglie wavelength =1.5$\pi {{a}_{0}}$
The equation can be written as:
\[2\pi {{a}_{0}}(\dfrac{{{n}^{2}}}{Z})=n(1.5\pi ){{a}_{0}}\]
The value of $\dfrac{n}{z}$= 0.75

Hence the correct answer is option (B) i.e. then the value of n/z is = 0.75

Additional information:
Let’s check the second postulate of Bohr i.e. he described the stable orbitals in his second postulate. And according to this postulate and electron found to be revolving around the nucleus of an atom. During this revolution the angular momentum is calculated as the integral multiple of $\dfrac{h}{2\pi }$. Where h is the planck's constant. If L is the angular momentum of the orbiting electron then L = $\dfrac{nh}{2\pi }$, where n is the principal quantum number of the given atom.

Note: Quantization of angular momentum is defined as the radius of the orbit which is found to be discrete which means that it is quantized. Not only the radius but the energy is also quantized.