Question

The wavelength of an electron revolving in the 3^rd orbit of He⁺ is ‘λ’. If the radius of first bohr orbit in H atom is ‘x’, then λ is equal to 
1. 6πx 
2. 3πx 
3. 4πx 
4. 9πx 

Answer 1

Correct Answer - Option 2 : 3πx 

Concept:

Bohr Model of the hydrogen atom first proposed the planetary model, but later an assumption concerning the electrons was made.

The assumption was the quantization of the structure of atoms. Bohr’s proposed that electrons orbited the nucleus in specific orbits or shells with a fixed radius. Only those shells with a radius provided by the equation below were allowed, and it was impossible for electrons to exist between these shells.

Mathematically, the allowed value of the atomic radius is given by the equation

r(n) = (n² / z)× r(1)

r(1) is the smallest allowed radius for the hydrogen atom also known as the Bohr’s radius.

r(1) = 0.529 A°;

According to bohr’s model, the angular momentum of electron is also quantized and is given by

\(mvr = \frac{{nh}}{{2\pi }}\)

Where n – level of the electron, r – radius of the n^th orbit, v – corresponding speed;

Calculation:

Given The wavelength of an electron revolving in the 3^rd orbit of He⁺ is ‘λ.

⇒ At n = 3, let r(3) = k

If the radius of first Bohr orbit in H atom is ‘x’ ⇒ r(1) = x;

Form the above relation,

r(3) = k = 4.5x;

Considering angular momentum equation,

\( \Rightarrow mvr = \frac{{nh}}{{2\pi }} \Rightarrow mv\left( {4.5x} \right) = \frac{{3h}}{{2\pi }}\)

\( \Rightarrow \frac{h}{{mv}} = 3\pi x\)

⇒ The wavelength of an electron revolving in the 3^rd orbit of He⁺ is λ = 3πx