Question

The de Broglie wave length of a riffle bullet of mass 2 grams moving with a velocity of 2m/sec
a) 1.65 x 10-34m
b ) 1.65 x 10-24 m c) 1.65 x 10-31m
d ) 1.65 x 10-27m​

Answer 1

Answer:

de broglie wavelength =h/p where h is planks constant and p is momentum

h/p=h/mv

2g =0.02kg

6.626 x 10 ^-34/.02kgx2m/s now

Explanation:

solve it

Answer 2

Answer:

The correct option is $C$ $1.65 * 10^{-31} m$.

Explanation:

The de Broglie equation exists an equation utilized to explain the wave properties of matter, especially, the wave nature of the electron:​

$&\lambda=\frac{h}{m v}$,

where λ exists wavelength, h exists Planck's constant, and m exists the mass of a particle, moving at a velocity v. de Broglie proposed that particles can display the effects of waves.

$&\frac{6.6 \times 10^{-34}}{2 \times 10^{-3} \times 2} m$

$&\lambda=\frac{h}{m v}$

$&\frac{6.625 \times 10^{-34}}{2 \times 10^{-3} \times 2}$

The correct option is $C$ $1.65 * 10^{-31} m$.

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