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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 21 of 67
Marks: +1, -0
The mass of an electron is 9.1 × 10−3110^{-31} kg. If its K.E. is 3.0 × 10−2510^{-25} J, calculate its wavelength.
Solution:  
K.E. = 12mv2\frac{1}{2} mv^2 , v = (2×3×10−253.1×10−31)12\left(\frac{2 \times 3 \times 10^{-25}}{3.1 \times 10^{-31}}\right)^{\frac{1}{2}} = 812 ms−1\mathrm{ms^{-1}}
Now, λ = hmv\frac{h}{mv} = 6.626×10−349.1×10−31×812\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 812} = 8.967 × 10−710^{-7} m = 8967 Å
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