Chapter 26, Problem 102QAP

To determine

(a)

The momentum and energy of photon.

Answer to Problem 102QAP

The Momentum of photon is $\text{1}\text{.326}\times {10}^{-25}\text{kg}\text{.m/s}$ and the energy of the photon is $3.978\times {10}^{-17}J$.

Explanation of Solution

Given info:

  $\text{Wavelength of microscope }\lambda =0.05nm$

Formula used:

  $\text{E=}\frac{hc}{\lambda }$

  $\text{momentum (p)=}\frac{h}{\lambda }$

De Broglie's wavelength

  $\lambda =\frac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}$

Calculation:

Substituting values in above equation

  $\text{momentum (p)=}\frac{h}{\lambda }\text{=}\frac{6.63\times {10}^{-34}}{0.05\times {10}^{-7}}\text{=1}\text{.326}\times {10}^{-25}\text{kg}\text{.m/s}$

  $\text{E=}\frac{hc}{\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{0.05\times {10}^{-7}}=3.978\times {10}^{-17}j$

Conclusion:

Thus, The Momentum of photon is $\text{1}\text{.326}\times {10}^{-25}\text{kg}\text{.m/s}$ and the energy of the photon is $3.978\times {10}^{-17}J$

To determine

(b)

The momentum and speed of electron

Answer to Problem 102QAP

The momentum of electron is $1.32\times {10}^{-24}kg.m/s$ and speed of electron is $1.45\times {10}^{6}m/s$ and kinetic energy of electron is $9.57\times {10}^{-19}\text{ J}$.

Explanation of Solution

Given:

  $\text{Wavelength of microscope }\lambda =0.05nm$

Calculation:

  $\begin{array}{l}\text{For De broglies wavelength, we know,}\\ \lambda =\frac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}\\ 0.05\times {10}^{-7}=\frac{6.63\times {10}^{-34}}{9.11\times {10}^{-31}v}\sqrt{1-\frac{v^2}{{(3\times {10}^8)}^2}}\\ v=1.45\times {10}^{6}m/s\\ p=mv=9.11\times {10}^{-31}\times 1.45\times {10}^6=1.32\times {10}^{-24}kg.m/s\\ KE=\frac{1}{2}m{v}^2=\frac{1}{2}[9.11\times {10}^{-31}\times {(}^{1.45}]=9.57\times {10}^{-19}\text{ J}\\ \end{array}$

Conclusion:

Thus, the momentum of electron is $1.32\times {10}^{-24}kg.m/s$ and speed of electron is $1.45\times {10}^{6}m/s$ and kinetic energy of electron is $9.57\times {10}^{-19}\text{ J}$.

To determine

(c)

The benefit of using an electron instead of photon.

Answer to Problem 102QAP

It is more beneficial to use an electron as the energy required for electron is less as compared to photon

Explanation of Solution

Introduction:

It is given that the $\text{Wavelength of microscope }\lambda =0.05nm$

Kinetic energy of electron is $9.57\times {10}^{-19}\text{ J}$

Energy of the photon is $3.978\times {10}^{-17}J$

As we can observe, energy required by an electron is less than that of photon.

Conclusion:

Thus, it is more beneficial to use an electron as the energy required for electron is less as compared to photon