Questions & Answers

Question

Answers

(A) 1.0

(B) 0.75

(C) 0.40

(D) 1.50

Answer

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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

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.

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