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Question

Chemistry Question on Structure of atom

The de-Broglie wavelength of a particle with mass 1 g and velocity 100 m/s is

A

6.63×10−33m6.63 \times 10^{-33} m

B

6.63×10−34m6.63 \times 10^{-34} m

C

6.63×10−35m6.63 \times 10^{-35} m

D

6.63×10−36m6.63 \times 10^{-36} m

Answer

6.63×10−33m6.63 \times 10^{-33} m

Explanation

Solution

p=hλp=\frac{h}{\lambda} (de-Broglie equation)
\lambda=\frac{h}{mv}\hspace15mm (\because p=mv)
h=6.625×10−34h = 6.625 \times 10^{-34}
≈6.63×10−34kg/s\approx 6.63 \times 10^{-34} kg/s
λ=6.63×10−34kgm2/s10−10kg×100m/s\lambda=\frac{6.63 \times10^{-34}kg m^2/ s}{10^{-10}kg \times 100m/s}
=6.63×10−33m=6.63 \times10^{-33} m