Question

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

Answer 1

According to de Broglie’s equation, $\lambda =\frac{h}{mv}$
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:
$\lambda =\frac{6.626\times {10}^{-34}Js}{(9.10939\times {10}^{-31}kg)(2.19\times {10}^{6}m{s}^{-1})}=3.32\times {10}^{-10}m=3.32\times {10}^{-10}m\times \frac{100}{100}=332\times {10}^{-12}m=\lambda =332pm$
$\therefore$ Wavelength associated with the electron = 332 pm