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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 59 of 67
Marks: +1, -0
If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 m s−110^{6}\,\mathrm{m}\,\mathrm{s}^{-1}, calculate the de Broglie wavelength associated with it.
Solution:  
We know that, λ = hmv\frac{h}{mv} where, h = 6.626 × 10−3410^{-34} J s,
m = mass of electron = 9.1 × 10−31 kg10^{-31}\,\mathrm{kg}, v = 2.19 × 106 m/s10^{6}\,\mathrm{m/s}
∴ λ = 6.626×10−349.1×10−31×2.19×106\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.19 \times 10^{6}} = 0.32 × 10−9 m10^{-9}\,\mathrm{m} = 332 × 10−12 m10^{-12}\,\mathrm{m}
λ = 332 pm [Since 1 pm = 10−12 m10^{-12}\,\mathrm{m}]
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