Question

If the Planck’s constant, $h=6.6\times {10}^{-34}$ Js, then the de Broglie wavelength of a particle in $\stackrel{\circ }{A}$ having momentum of $3.3\times {10}^{-24}$ kg m $s^s$ will be :

• A. 0.02
• B. 0.5
• C. 2
• D. 500

Answer 1

The correct option is C 2

$\lambda =\frac{h}{mv}=\frac{6.6\times {10}^{-34}}{\mathrm{3.3.}\times {10}^{-24}}=2\times {10}^{-10}m=2\stackrel{\circ }{A}$