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Which is the correct relation between de-Broglie wavelength of an electron in the nth Bohr orbit and radius of the orbit R?
1. λ = n2πR 
2. \(\lambda = \dfrac{2\pi R}{n}\)
3. \(\lambda = \dfrac{4\pi R}{n}\)
4. \(\lambda = \dfrac{2\pi R}{nh}\)

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Correct Answer - Option 2 : \(\lambda = \dfrac{2\pi R}{n}\)

Concept: 

The angular momentum of an electron in a given stationary state can be expressed as

\(m_e v R = \dfrac{nh}{2\pi}\).......................(1), n = 1,2,3.....

where R = radius of the orbit, me= mass of electron, v = velocity of electron h = Planck constant

Thus an electron can move only in those orbits for which its angular momentum is an integral multiple of h/2π that is why only certain fixed orbits are allowed.

According to Bohr’s theory for hydrogen atom:

a) The stationary states for electron are numbered n = 1,2,3.......... These integral numbers are known as Principal quantum numbers.

b) The radii of the stationary states are expressed as where a0 = 52,9 pm.

Thus the radius of the first stationary state called the Bohr orbit

The de-Broglie gives the relation between wavelength and momentum in other words wave and particle nature of the electron. 

The de-Broglie wavelength is given as,

\(\lambda = \dfrac{h}{p} = \dfrac{h}{m_ev}\)......................(2)

here p = momentum of electron

 

Calculation:

To find: the correct relationship between de-Broglie wavelength and radius of the electron

From equation 1 we have,

\(m_e v r = \dfrac{nh}{2\pi}\)

\(\dfrac{h}{m_ev} =\dfrac{ 2\pi R}{n} \)

From equation 2 we have 

\(\lambda = \dfrac{h}{m_ev}\)

On comparing both the relation we get

\(\lambda = \dfrac{2\pi R}{n}\)

Hence the correct option is 2

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