#### Question

If the velocity of the electron in Bohr’s first orbit is 2.19 × 10^{6} ms^{–1}, calculate the de Broglie wavelength associated with it.

#### Solution

According to de Broglie’s equation,

`lambda = h/(mv)`

Where,

λ = wavelength associated with the electron

*h* = Planck’s constant

*m* = mass of electron

*v* = velocity of electron

Substituting the values in the expression of *λ**:*

`lambda = (6.626 xx 10^(-34) Js)/((9.10939 xx 10^(-31)kg)(2.19xx10^6 ms^(-1))`

`= 3.32 xx 10^(-10)m = 3.32 xx 10^(-10) mxx 100/100`

`= 332xx10^(-12) m`

λ = 332 pm

∴Wavelength associated with the electron = 332 pm

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Solution If the Velocity of the Electron in Bohr’S First Orbit is 2.19 × 106 Ms–1, Calculate the De Broglie Wavelength Associated with It. Concept: Bohr'S Model for Hydrogen Atom.