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Question: If the velocity of an electron in Bohr’s first orbit is \(2.19 \times 10^{6}ms^{- 1}\) , what will b...

If the velocity of an electron in Bohr’s first orbit is 2.19×106ms−12.19 \times 10^{6}ms^{- 1} , what will be the de Broglie wavelength associated with it?

A

2.19×10−6m2.19 \times 10^{- 6}m

B

4.38×10−6m4.38 \times 10^{- 6}m

C

3.32×10−10m3.32 \times 10^{- 10}m

D

3.32×1010m3.32 \times 10^{10}m

Answer

3.32×10−10m3.32 \times 10^{- 10}m

Explanation

Solution

: λ=hmv=6.626×10−34Js9.11×10−31kg×2.19×106ms−1\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{- 34}Js}{9.11 \times 10^{- 31}kg \times 2.19 \times 10^{6}ms^{- 1}}

=3.32×10−10m= 3.32 \times 10^{- 10}m