Chapter 2, Problem 2.15P

(a)

To determine

The voltage required to accelerate the proton.

Answer to Problem 2.15P

The voltage required to accelerate the proton from zero velocity to a velocity corresponding to a de Broglie wavelength of $0.01\text{A}$ is $816\text{eV}$ .

Explanation of Solution

Given:

De-Broglie Wavelength, $\lambda =0.01\text{A}=1\times {10}^{-12}\text{m}$

Mass of the proton, $m_p=1.67\times {10}^{-27}\text{kg}$

Initial velocity, $v_i=0\text{m/s}$

Formula used:

The de-Broglie wavelength of the particle is given as

  $\lambda =\frac{h}{mv}$

Here,

  $h=6.64\times {10}^{-34}\text{J-s}$ is Planck’s constant

  $m$ is a mass of the particle

  $v$ is the velocity of the particle

Calculation:

Calculating the velocity of the proton corresponding to a wavelength of $0.01\text{A}$ , considering the relativistic effect

  $\lambda =\frac{h}{mv}$

Plugging the values in the above equation:

  $\begin{array}{c}1\times {10}^{-12}=\frac{6.64\times {10}^{-34}}{mv}\\ mv=6.63\times {10}^{-22}\text{kg-m/s}\end{array}$

We need to check whether the relativistic effect is important

  $\begin{array}{c}p=mv=\frac{m_o}{\sqrt{1-{\beta }^2}}\beta c\\ 6.63\times {10}^{-22}=\frac{1.627\times {10}^{-27}\times \beta \times 3\times {10}^8}{\sqrt{1-{\beta }^2}}\\ \beta =\frac{v}{c}=1.32\times {10}^{-3}<0.1\\ v=3.95\times {10}^5\text{m/s}\end{array}$

The voltage required to accelerate the proton from zero velocity to a velocity corresponding to a de Broglie wavelength of $0.01\text{A}$ is calculated as

  $\begin{array}{c}\Delta PE=\Delta KE\\ qV=\frac{1}{2}m\left(v_f^2-v_i^2\right)\\ 1.6\times {10}^{-19}\times V=\frac{1}{2}\times 1.67\times {10}^{-27}\left({\left(3.95\times {10}^5\right)}^2-0^2\right)\\ V=816\text{eV}\end{array}$

Conclusion:

Thus,the voltage required to accelerate the proton from zero velocity to a velocity corresponding to a de Broglie wavelength of $0.01\text{A}$ is $816\text{eV}$ .

(b)

To determine

The kinetic energy of an electron corresponding to a de Broglie wavelength of $0.01\text{A}$ .

Answer to Problem 2.15P

The kinetic energy of an electron corresponding to a de Broglie wavelength of $0.01\text{A}$ is $0.83\text{MeV}$ .

Explanation of Solution

Given:

De-Broglie Wavelength, $\lambda =0.01\text{A}=1\times {10}^{-12}\text{m}$

Mass of the electron, $m_e=9.1\times {10}^{-31}\text{kg}$

Formula used:

Kinetic energy is given as

  $KE=\frac{1}{2}m{v}^2$

Here,

  $m$ is a mass of the particle

  $v$ is the velocity of the particle

Calculation:

The velocity of the electron corresponding to a wavelength of $0.01\text{A}$

  $\lambda =\frac{h}{mv}$

Plugging the values in the above equation

  $\begin{array}{c}1\times {10}^{-12}=\frac{6.64\times {10}^{-34}}{9.1\times {10}^{-31}\times v}\\ v=7.296\times {10}^8\text{m/s}>3\times {10}^8\end{array}$

Since the speed of the electron is greater than the speed of light in a vacuum, therefore, we need to consider the effect of relativity

We need to check whether the relativistic effect is important

  $\begin{array}{c}p=mv=\frac{m_o}{\sqrt{1-{\beta }^2}}\beta c\\ 6.63\times {10}^{-22}=\frac{9.11\times {10}^{-31}\times \beta \times 3\times {10}^8}{\sqrt{1-{\beta }^2}}\\ \beta =\frac{v}{c}=0.92>0.1\\ v=2.76\times {10}^8\text{m/s}\end{array}$

The kinetic energy of the electron corresponding to a wavelength of $0.01\text{A}$ is calculated as:

  $\begin{array}{c}\Delta KE=m_o{c}^2\left(\frac{1}{\sqrt{1-{\beta }^2}}-1\right)\\ =9.1\times {10}^{-31}\times {\left(3\times {10}^8\right)}^2\left(\frac{1}{\sqrt{1-{0.92}^2}}-1\right)\\ =1.27\times {10}^{-13}\text{J}\\ =\frac{1.27\times {10}^{-13}}{1.6\times {10}^{-19}}\left(\text{eV}\right)\\ =0.83\times {10}^6\text{eV}\\ =0.83\text{MeV}\end{array}$

Conclusion:

The kinetic energy of an electron corresponding to a de Broglie wavelength of $0.01\text{A}$ is $0.83\text{MeV}$ .

(c)

To determine

The energy of an X-ray photon corresponding to a wavelength of $0.01\text{A}\text{.}$

Answer to Problem 2.15P

The energy of an X-ray photon corresponding to a wavelength of $0.01\text{A}$ is $\text{1}\text{.245}\text{MeV}$ .

Explanation of Solution

Given:

Wavelength, $\lambda =0.01\text{A}=1\times {10}^{-12}\text{m}$

Formula used:

The energy of a photon is given as:

  $E=\frac{hc}{\lambda }$

Here,

  $h$ is Planck’s constant

  $c$ is the speed of light

  $\lambda$ is the wavelength of the photon

Calculation:

The energy of a photon is given as

  $E=\frac{hc}{\lambda }$

Plugging the values in the above equation

  $\begin{array}{c}E=\frac{6.64\times {10}^{-34}\times 3\times {10}^8}{1\times {10}^{-12}}\\ =1.992\times {10}^{-13}\text{J}\\ =\frac{1.992\times {10}^{-13}}{1.6\times {10}^{-19}}\left(\text{eV}\right)\\ =1.245\times {10}^6\text{eV}\\ =\text{1}\text{.245}\text{MeV}\end{array}$

Conclusion:

The energy of an X-ray photon corresponding to a wavelength of $0.01\text{A}$ is $\text{1}\text{.245}\text{MeV}$ .