De Broglie wave length of electrons

Worked out. Three everyday-to-atomic-scale objects are compared: (i) an electron of kinetic energy 2 eV, (ii) a 50 g rifle bullet moving at 200 m/s, and (iii) a 4000 kg car moving at 50 m/s on a highway. For the electron, its momentum is found from $p=\sqrt{2mK}\approx7.63\times10^{-25}$ kg m/s, giving a de Broglie wavelength of $\lambda=h/p\approx8.68$ Å -- comparable to atomic spacings and easily observable via diffraction. For the bullet, momentum is simply $p=mv=0.050\times200=10$ kg m/s, giving an utterly negligible $\lambda\approx6.626\times10^{-35}$ m ($\approx10^{-33}$ m scale once compared against measurement limits). For the car, $p=mv=4000\times50=2\times10^5$ kg m/s, giving an even smaller $\lambda\approx3.313\times10^{-39}$ m. The three results together make the chapter's key point concrete: matter's wave nature is significant and measurable at the atomic/subatomic scale but is utterly irrelevant -- far b …

Worked out. An alpha particle (2 protons + 2 neutrons, so mass $M=4m_p$ and charge $q=2e$) is accelerated through a potential difference of 400 V, using the proton mass $m_p=1.67\times10^{-27}$ kg, and the task is to find its de Broglie wavelength. Using the accelerated-charged-particle form $\lambda=h/\sqrt{2MqV}=h/\sqrt{2(4m_p)(2e)V}$ and substituting all the given numbers gives $\lambda\approx0.00507$ Å -- roughly two thousand times shorter than the 100 V electron's 1.227 Å wavelength worked out earlier in this section, because the alpha particle's much larger mass and charge both act to shrink its de Broglie wavelength for a comp …

Worked out. A proton and an electron happen to share the exact same de Broglie wavelength, and the task is to determine which one moves faster and which one possesses the greater kinetic energy. Starting from $\lambda=h/\sqrt{2mK}$, equating the two particles' wavelengths gives $K_p/K_e=m_e/m_p$; since the electron's mass $m_e$ is thousands of times smaller than the proton's mass $m_p$, this ratio is far less than one, meaning the electron carries the greater kinetic energy. Separately, writing $K=\tfrac12mv^2$ and again equating wavelengths gives $v_p/v_e=m_e/m_p$, which is likewise far less than one, meaning the electron also moves faster than the proton. The example's teaching point is that for two particles sharing an identical de Broglie wavelength, the lighter particle (the electron) is both the faster-moving one and the one with m …