Let \( \lambda_e \), \( \lambda_p \), and \( \lambda_d \) be the wavelengths associated with an electron, a proton, and a deuteron, all moving with the same speed. Then the correct relation between them is:
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The de Broglie wavelength is larger for lighter particles when moving at the same speed. Always compare the masses to determine the correct wavelength order.
The de Broglie wavelength (\(\lambda\)) of a particle is calculated using the formula: \[ \lambda = \frac{h}{m v} \] In this equation, \( h \) represents Planck’s constant, \( m \) is the particle's mass, and \( v \) is its velocity. Given that all three particles (electron, proton, and deuteron) share the same speed, the wavelength is inversely proportional to their respective masses: \[ \lambda \propto \frac{1}{m} \] The masses of these particles are ordered as follows: \( m_e \) (electron) is the least massive, \( m_p \) (proton) is more massive than the electron, and \( m_d \) (deuteron) is the most massive. Consequently, their de Broglie wavelengths are ordered in the reverse manner: \[ \lambda_e > \lambda_p > \lambda_d \] Therefore, the correct answer is (2).
