Dawn of Modern Physics Numerical Problems - Chapter 19 of FSc/ICS Physics

Dawn of Modern Physics Numerical Problems - Chapter 19

Problem 19.1: A particle called the pion lives on the average only about \(2.6 \times 10^{-8} \, \text{s}\) when at rest in the laboratory. It then changes to another form. How long would such a particle live when shooting through the space at 0.95c?

Given Data:

• Proper Time \(t_0 = 2.6 \times 10^{-8} \, \text{s}\)
• Relative Speed \(v = 0.95c\)

To Determine:

• Relativistic Time \(t = ?\)

Calculations:

\[ t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{2.6 \times 10^{-8}}{\sqrt{1 - (0.95)^2}} = \frac{2.6 \times 10^{-8}}{\sqrt{1 - 0.9025}} = \frac{2.6 \times 10^{-8}}{\sqrt{0.0975}} \] \[ t = \frac{2.6 \times 10^{-8}}{0.3122} = 8.33 \times 10^{-8} \, \text{s} \]

Problem 19.2: What is the mass of a 70 kg man in a space rocket traveling at 0.8c from us as measured from Earth?

Given Data:

• Proper Mass \(m_0 = 70 \, \text{kg}\)
• Relative Speed \(v = 0.8c\)

To Determine:

• Relativistic Mass \(m = ?\)

Calculations:

\[ m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{70}{\sqrt{1 - (0.8)^2}} = \frac{70}{\sqrt{1 - 0.64}} = \frac{70}{\sqrt{0.36}} = \frac{70}{0.6} = 116.67 \, \text{kg} \]

Problem 19.3: Find the energy of photon in (a) Radiowave of wavelength 100 m (b) Green light of wavelength 550 nm (c) X-ray with wavelength 0.2 nm

Given Data:

• (a) Wavelength of Radiowave \(\lambda_1 = 100 \, \text{m}\)
• (b) Wavelength of Green Light \(\lambda_2 = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m}\)
• (c) Wavelength of X-ray \(\lambda_3 = 0.2 \, \text{nm} = 0.2 \times 10^{-9} \, \text{m}\)

To Determine:

• (a) Energy of Radiowave \(E_1 = ?\)
• (b) Energy of Green Light \(E_2 = ?\)
• (c) Energy of X-ray \(E_3 = ?\)

Calculations:

(a) Energy of Radiowave:

\[ E_1 = \frac{hc}{\lambda_1} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{100} = 1.989 \times 10^{-27} \, \text{J} \] \[ E_1 = \frac{1.989 \times 10^{-27}}{1.6 \times 10^{-19}} = 1.243 \times 10^{-8} \, \text{eV} \]

(b) Energy of Green Light:

\[ E_2 = \frac{hc}{\lambda_2} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}} = 3.616 \times 10^{-19} \, \text{J} \] \[ E_2 = \frac{3.616 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.26 \, \text{eV} \]

(c) Energy of X-ray:

\[ E_3 = \frac{hc}{\lambda_3} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.2 \times 10^{-9}} = 9.945 \times 10^{-16} \, \text{J} \] \[ E_3 = \frac{9.945 \times 10^{-16}}{1.6 \times 10^{-19}} = 6215.6 \, \text{eV} \]

Problem 19.4: Yellow light of 577 nm wavelength is incident on a cesium surface. The stopping voltage is found to be 0.25 V. Find (a) the Maximum K.E. of the photoelectrons (b) the work function of cesium

Given Data:

• Wavelength of Yellow Light \(\lambda = 577 \, \text{nm} = 577 \times 10^{-9} \, \text{m}\)
• Stopping Potential \(V_0 = 0.25 \, \text{V}\)

To Determine:

• (a) Maximum K.E. \(K.E._{\text{max}} = ?\)
• (b) Work Function \(\phi = ?\)

Calculations:

(a) Maximum Kinetic Energy:

\[ K.E._{\text{max}} = eV_0 = 1.6 \times 10^{-19} \times 0.25 = 4 \times 10^{-20} \, \text{J} \]

(b) Work Function:

\[ \phi = hf - K.E._{\text{max}} = \frac{hc}{\lambda} - K.E._{\text{max}} \] \[ \phi = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{577 \times 10^{-9}} - 4 \times 10^{-20} = 3.447 \times 10^{-19} - 0.04 \times 10^{-18} = 3.047 \times 10^{-19} \, \text{J} \] \[ \phi = \frac{3.047 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.904 \, \text{eV} \]

Problem 19.5: X-rays of wavelength 22 pm are scattered from a carbon target. The scattered radiation being viewed at 85° to the incident beam. What is Compton shift?

Given Data:

• Wavelength \(\lambda = 22 \, \text{pm} = 22 \times 10^{-12} \, \text{m}\)
• Scattering Angle \(\theta = 85^\circ\)

To Determine:

• Compton Shift \(\Delta\lambda = ?\)

Calculations:

\[ \Delta\lambda = \frac{h}{m_e c} (1 - \cos\theta) = 2.426 \times 10^{-12} \times (1 - \cos 85^\circ) \] \[ \Delta\lambda = 2.426 \times 10^{-12} \times (1 - 0.0872) = 2.426 \times 10^{-12} \times 0.9128 = 2.214 \times 10^{-12} \, \text{m} = 2.214 \, \text{pm} \]

Problem 19.6: A 90 keV X-ray photon is fired at a carbon target and Compton scattering occurs. Find the wavelength of the incident photon and the wavelength of the scattered photon for scattering angle of (a) 30° (b) 60°

Given Data:

• Energy of Photon \(E = 90 \, \text{keV} = 90 \times 10^3 \times 1.6 \times 10^{-19} \, \text{J} = 1.44 \times 10^{-14} \, \text{J}\)

To Determine:

• (a) Wavelength of Incident Photon \(\lambda = ?\)
• (b) Wavelength of Scattered Photon at 30° \(\lambda' = ?\)
• (c) Wavelength of Scattered Photon at 60° \(\lambda'' = ?\)

Calculations:

(a) Wavelength of Incident Photon:

\[ E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.44 \times 10^{-14}} = 13.81 \times 10^{-12} \, \text{m} = 13.81 \, \text{pm} \]

(b) Wavelength of Scattered Photon at 30°:

\[ \Delta\lambda = \frac{h}{m_e c} (1 - \cos\theta) = 2.426 \times 10^{-12} \times (1 - \cos 30^\circ) \] \[ \Delta\lambda = 2.426 \times 10^{-12} \times (1 - 0.8660) = 2.426 \times 10^{-12} \times 0.1340 = 0.325 \times 10^{-12} \, \text{m} \] \[ \lambda' = \lambda + \Delta\lambda = 13.81 \times 10^{-12} + 0.325 \times 10^{-12} = 14.135 \times 10^{-12} \, \text{m} = 14.135 \, \text{pm} \]

(c) Wavelength of Scattered Photon at 60°:

\[ \Delta\lambda = \frac{h}{m_e c} (1 - \cos\theta) = 2.426 \times 10^{-12} \times (1 - \cos 60^\circ) \] \[ \Delta\lambda = 2.426 \times 10^{-12} \times (1 - 0.5) = 2.426 \times 10^{-12} \times 0.5 = 1.213 \times 10^{-12} \, \text{m} \] \[ \lambda'' = \lambda + \Delta\lambda = 13.81 \times 10^{-12} + 1.213 \times 10^{-12} = 15.023 \times 10^{-12} \, \text{m} = 15.023 \, \text{pm} \]

Problem 19.7: What is the maximum wavelength of the two photons produced when a positron annihilates an electron? The rest mass energy of each is 0.51 MeV.

Given Data:

• Rest Mass Energy \(E = 0.51 \, \text{MeV} = 0.51 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J} = 8.16 \times 10^{-14} \, \text{J}\)

To Determine:

• Wavelength \(\lambda = ?\)

Calculations:

\[ E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{8.16 \times 10^{-14}} = 2.438 \times 10^{-12} \, \text{m} = 2.438 \, \text{pm} \]

Problem 19.8: Calculate the wavelength of (a) a 140 g ball moving at 40 ms⁻¹ (b) a proton moving at the same speed (c) an electron moving at the same speed

Given Data:

• Speed \(v = 40 \, \text{ms}^{-1}\)
• (a) Mass of Ball \(m_1 = 140 \, \text{g} = 0.14 \, \text{kg}\)
• (b) Mass of Proton \(m_2 = 1.67 \times 10^{-27} \, \text{kg}\)
• (c) Mass of Electron \(m_3 = 9.1 \times 10^{-31} \, \text{kg}\)

To Determine:

• (a) Wavelength associated with Ball \(\lambda_1 = ?\)
• (b) Wavelength associated with Proton \(\lambda_2 = ?\)
• (c) Wavelength associated with Electron \(\lambda_3 = ?\)

Calculations:

(a) Wavelength of Ball:

\[ \lambda_1 = \frac{h}{m_1 v} = \frac{6.63 \times 10^{-34}}{0.14 \times 40} = 1.183 \times 10^{-34} \, \text{m} \]

(b) Wavelength of Proton:

\[ \lambda_2 = \frac{h}{m_2 v} = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 40} = 9.92 \times 10^{-9} \, \text{m} = 9.92 \, \text{nm} \]

(c) Wavelength of Electron:

\[ \lambda_3 = \frac{h}{m_3 v} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 40} = 1.82 \times 10^{-5} \, \text{m} \]

Problem 19.9: What is the de Broglie wavelength of an electron whose kinetic energy is 120 eV?

Given Data:

• Kinetic Energy \(K.E. = 120 \, \text{eV} = 120 \times 1.6 \times 10^{-19} \, \text{J} = 1.92 \times 10^{-17} \, \text{J}\)
• Mass of Electron \(m = 9.1 \times 10^{-31} \, \text{kg}\)

To Determine:

• De Broglie Wavelength \(\lambda = ?\)

Calculations:

\[ K.E. = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2 \times K.E.}{m}} = \sqrt{\frac{2 \times 1.92 \times 10^{-17}}{9.1 \times 10^{-31}}} = 6.5 \times 10^6 \, \text{ms}^{-1} \] \[ \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 6.5 \times 10^6} = 1.12 \times 10^{-10} \, \text{m} \]

Problem 19.10: An electron is placed in a box about the size of an atom that is about \(1.0 \times 10^{-10} \, \text{m}\). What is the velocity of the electron?

Given Data:

• Size of Box \(\Delta x = 1.0 \times 10^{-10} \, \text{m}\)
• Mass of Electron \(m = 9.1 \times 10^{-31} \, \text{kg}\)

To Determine:

• Velocity \(\Delta v = ?\)

Calculations:

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \Rightarrow \Delta x \cdot m\Delta v \geq \frac{h}{4\pi} \] \[ \Delta v \geq \frac{h}{4\pi m \Delta x} = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 1.0 \times 10^{-10}} = 5.79 \times 10^5 \, \text{ms}^{-1} \]