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.

Answer 1

Answer

According to de Broglie’s equation,

λ = 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:

λ = 6.626x10-34 / (9.11x10-31)(2.19x106)

=3.32x10-10

= 332 pm

Answer 2

Explanation:

According to de Broglie's equation

λ=

mv

h

where, λ= de-Broglie wavelength

h=6.626×10

−34

Js=Planck's constant

m=mass of electron=9.1×10

−31

kg

velocity of electron, v=2.19×10

6

m/s

upon substitution we get:

λ=

9.1×10

−31

×2.19×10

6

6.626×10

−34

λ=0.332×10

−9

=3.32×10

−10

m