Question

The kinetic energies of an electron (\(e\)) and a proton (\(p\)) are \(E\) and \(3E\), respectively. Given that the mass of a proton is 1836 times that of an electron, the ratio of their de Broglie wavelengths (\(\lambda_e / \lambda_p\)) is ___________ (rounded off to two decimal places).

Hint:

% Quick Tip The de Broglie wavelength is inversely proportional to the momentum of the particle. For particles with the same kinetic energy, the particle with the larger mass will have a smaller de Broglie wavelength. Conversely, for particles with the same momentum, the particle with the larger mass will have a smaller kinetic energy.

Answer 1

Correct Answer: 74.1

The de Broglie wavelength (\(\lambda\)) of a particle is given by the equation: \[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant and \(p\) is the momentum of the particle. The kinetic energy (\(K\)) of a particle is related to its momentum by: \[ K = \frac{p^2}{2m} \implies p = \sqrt{2mK} \] where \(m\) is the mass of the particle. For the electron (\(e\)), the kinetic energy is \(K_e = E\) and its mass is \(m_e\). The de Broglie wavelength of the electron is: \[ \lambda_e = \frac{h}{p_e} = \frac{h}{\sqrt{2m_e K_e}} = \frac{h}{\sqrt{2m_e E}} \] For the proton (\(p\)), the kinetic energy is \(K_p = 3E\) and its mass is \(m_p = 1836 m_e\). The de Broglie wavelength of the proton is: \[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p K_p}} = \frac{h}{\sqrt{2(1836 m_e) (3E)}} = \frac{h}{\sqrt{11016 m_e E}} \] Now, we need to find the ratio of their de Broglie wavelengths (\(\lambda_e / \lambda_p\)): \[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_e E}}}{\frac{h}{\sqrt{11016 m_e E}}} = \frac{\sqrt{11016 m_e E}}{\sqrt{2m_e E}} = \sqrt{\frac{11016 m_e E}{2m_e E}} \] \[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{11016}{2}} = \sqrt{5508} \] \[ \sqrt{5508} \approx 74.2159 \] Rounding off to two decimal places, the ratio \(\lambda_e / \lambda_p\) is 74.22. This falls within the given range of 74.10 to 74.30.