Question

By means of the diffraction experiment, it is determined that the electron's de Broglie wavelength is $6.6\times {10}^{-10}m$. What is the electron's linear momentum? Use Planck's constant, $h=6.6\times {10}^{-34}JAs$.

• A. $1.0\times {10}^{-44}kgAm/s$
• B. $1.0\times {10}^{-24}kgAm/s$
• C. $1.0\times {10}^{24}kgAm/s$
• D. $2.0\times {10}^{24}kgAm/s$
• E. $1.0\times {10}^{44}kgAm/s$

Answer 1

The correct option is B $1.0\times {10}^{-24}kgAm/s$

From de Broglie hypothesis , the linear momentum, $p=\frac{h}{\lambda }=\frac{6.6\times {10}^{-34}}{6.6\times {10}^{-10}}={10}^{-24}kg{A}^{o}m/s$