91Ó°ÊÓ

To evaluate the wavelength, of the value λ, of a photon with a given momentum. Consider the equation of the wavelength:

\[{\bf{\lambda = }}\frac{{\bf{h}}}{{\bf{\rho }}}\]

Consider the formula for the momentum is:

\[{\bf{p}} = {\bf{mv}}\]

Considering the given information:

\[\begin{array}{c}h = 6.63 \times {10^{ - 34}}\;{\rm{J}}{{\rm{s}}^{ - 1}}\\c \approx 3.0 \times {10^8}\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}\end{array}\]

Derive the formula for the momentum as:

\[p \approx \frac{h}{\lambda }\]

Or

\[\lambda \approx \frac{h}{p}\]

Or

\[p \approx mv\]

Consider the velocity of an electron moves at 3% of the speed of light.

\[\begin{array}{l}v \approx 3\% \times c\\v \approx \frac{3}{{100}} \times 3 \times {10^8}\\v \approx 9 \times {10^6}\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\]

Substitute and determine the momentum.

\[\begin{array}{l}p \approx 9.11 \times {10^{ - 31}} \times 9 \times {10^6}\\p \approx 8.2 \times {10^{ - 24}}\;\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\end{array}\]

Currently, the electron's velocity is:

\[\begin{array}{l}\lambda \approx \frac{{6.67 \times {{10}^{ - 34}}}}{{8.2 \times {{10}^{ - 24}}}}\\v \approx 8.07 \times {10^{ - 11}}\;{\rm{m}}\end{array}\]

Therefore, the required wavelength of electron is\[8.07 \times {10^{ - 11}}\;{\rm{m}}\].