When an electron is accelerated through a potential difference \( V \), its kinetic energy is:
\[ KE = eV \]
Given:
Using the de Broglie equation:
\[ \lambda = \frac{12.26}{\sqrt{V}} Å \]
\[ \lambda = \frac{12.26}{\sqrt{150}} Å \]
\[ \lambda = \frac{12.26}{12.24} Å \]
\[ \lambda = 1.001 Å \]
Energy of a Photon
The energy of a photon is given by the equation:
\[ E_p = \frac{hc}{\lambda} \]where:
Substituting the values:
\[ E_p = \frac{(6.626 \times 10^{-34}) (3.0 \times 10^8)}{10^{-10}} \] \[ E_p = 19.878 \times 10^{-16} \text{ J} \]Energy of a Neutron
The energy of a neutron (de Broglie relation) is given by:
\[ E_n = \frac{p^2}{2m}, \quad \text{where} \quad p = \frac{h}{\lambda} \]Substituting \( p \) into the equation:
\[ E_n = \frac{(h/\lambda)^2}{2m} \]Given:
Substituting values:
\[ E_n = \frac{(6.626 \times 10^{-34})^2}{2 (1.67 \times 10^{-27}) (10^{-10})^2} \] \[ E_n = 13.144 \times 10^{-21} \text{ J} \]Step 3: Compare the Energies
Taking the ratio:
\[ \frac{E_p}{E_n} = \frac{19.878 \times 10^{-16}}{13.144 \times 10^{-21}} \] \[ \frac{E_p}{E_n} \approx 1.512 \times 10^5 \]Conclusion:
The energy of a photon is approximately \( 1.512 \times 10^5 \) times greater than the energy of a neutron when both have the same wavelength of 1Å.
For an electron:
\[ \lambda_e = \frac{h}{\sqrt{2m_e eV}} \]
For a proton:
\[ \lambda_p = \frac{h}{\sqrt{2m_p eV}} \]
Taking the ratio of the two wavelengths:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_p}}{\sqrt{m_e}} \]
Given:
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{1.673 \times 10^{-27}}{9.109 \times 10^{-31}}} \]
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{1836} \]
\[ \frac{\lambda_e}{\lambda_p} \approx 42.85 \]
The electron has a de Broglie wavelength approximately 43 times larger than that of the proton when accelerated through the same potential difference \( V \).
The de Broglie wavelength equation is:
\[ \lambda = \frac{h}{p} \]
where \( \lambda = 5000 \) Å \( = 5.0 \times 10^{-7} \) m, and \( h = 6.626 \times 10^{-34} \) J·s.
\[ p = \frac{h}{\lambda} \]
\[ p = \frac{6.626 \times 10^{-34}}{5.0 \times 10^{-7}} \]
\[ p = 1.325 \times 10^{-27} \text{ kg·m/s} \]
The kinetic energy formula is:
\[ KE = \frac{p^2}{2m} \]
where \( m_e = 9.109 \times 10^{-31} \) kg.
\[ KE = \frac{(1.325 \times 10^{-27})^2}{2 \times (9.109 \times 10^{-31})} \]
\[ KE = \frac{1.756 \times 10^{-54}}{1.822 \times 10^{-30}} \]
\[ KE = 9.64 \times 10^{-25} \text{ J} \]
Since \( 1 eV = 1.602 \times 10^{-19} \) J:
\[ KE = \frac{9.64 \times 10^{-25}}{1.602 \times 10^{-19}} \]
\[ KE \approx 6.02 \times 10^{-6} \text{ eV} \]
The de Broglie wavelength is related to momentum by: \[ \lambda = \frac{h}{p} \] where \( \lambda = 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \) and \( h = 6.626 \times 10^{-34} \, \text{J·s} \).
The momentum of the neutron is: \[ p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{1 \times 10^{-10}} = 6.626 \times 10^{-24} \, \text{kg·m/s} \]
The kinetic energy of the neutron is given by: \[ K.E. = \frac{p^2}{2m}\] where \( m = 1.675 \times 10^{-27} \, \text{kg} \).
Substituting the values: \[ K.E. = \frac{(6.626 \times 10^{-24})^2}{2 \times 1.675 \times 10^{-27}} \approx 1.309 \times 10^{-22} \, \text{J} \]
Finally, converting the energy to electron volts: \[ K.E. = \frac{1.309 \times 10^{-22}}{1.602 \times 10^{-19}} \approx 0.082 \, \text{eV} \]
Problems on the De-Broglie Hypothesis help students apply wave-particle duality to real-world cases of electrons, protons, and neutrons. By practicing numerical examples, learners understand how to calculate De-Broglie wavelength using kinetic energy, momentum, and potential difference.