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

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$2.19 \times 10^{- 6} m$

$4.38 \times 10^{- 6} m$

$3.32 \times 10^{- 10} m$

$3.32 \times 10^{10} m$

answer is C.

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Detailed Solution

$\begin{array}{l} λ = \frac{h}{mv} = \frac{6 .626 \times 10^{- 34} Js}{9 .11 \times 10^{- 31} kg \times 2 .19 \times 10^{6} {ms}^{- 1}} \\ = 3 .32 \times 10^{- 10} m \end{array}$

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