Question
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.
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.
