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