91Ó°ÊÓ

To evaluate the wavelength, of the valueλ, 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:

\[\begin{array}{l}h = 6.63 \times {10^{ - 34}}\;{\rm{J}}{{\rm{s}}^{ - 1}}\\\lambda \approx 1.0\;\;{\rm{m}}\end{array}\]

Apply the formula:

The relationship between momentum and wavelength in electrons is :

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

Substitute the values and determine the electron momentum.

\[\begin{array}{l}p \approx \frac{{6.63 \times {{10}^{ - 34}}\;{\rm{J}}{{\rm{s}}^{ - 1}}}}{{1\;\;{\rm{m}}}}\\p \approx 6.63 \times {10^{ - 34}}\;\;{\rm{kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - 1}}\end{array}\]

Currently, the electron's velocity is obtained as follows:

\[\begin{array}{l}v \approx \frac{p}{m}\\v \approx \frac{{6.63 \times {{10}^{ - 34}}}}{{9.11 \times {{10}^{ - 31}}\;{\rm{kg}}}}\\v \approx 7.27 \times {10^{ - 4}}\;\;{\rm{m}}{{\rm{s}}^{ - 1}}\end{array}\]

Therefore,the required velocity of electron is\[7.27 \times {10^{ - 4}}\;\;{\rm{m}}{{\rm{s}}^{ - 1}}\].