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Mass of particle from de Broglie ratio

Short answer

From λ = h/(mv), λ_p/λ_e = (m_e v_e)/(m_p v_p). With v_p = 3v_e, m_p = m_e/(3 × ratio) = 9.109×10⁻³¹/(3×1.813×10⁻⁴) ≈ 1.675×10⁻²⁷ kg.

A non relativistic particle is moving three times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.813 × 10⁻⁴. Calculate the mass of the particle.

de Broglie wavelength: λ = h/(mv). For the particle and the electron:

λ_p/λ_e = (m_e v_e)/(m_p v_p)

Given: v_p = 3 v_e, λ_p/λ_e = 1.813×10⁻⁴, m_e = 9.109×10⁻³¹ kg.

λ_p/λ_e = m_e v_e / (m × 3 v_e) = m_e / (3m)

m = m_e / (3 × λ_p/λ_e)
  = 9.109×10⁻³¹ / (3 × 1.813×10⁻⁴)
  = 9.109×10⁻³¹ / 5.439×10⁻⁴
  = 1.675×10⁻²⁷ kg

Answer: m ≈ 1.675×10⁻²⁷ kg, which is the mass of a neutron (about 1839 m_e).

✓ Answer checked by Claude Opus 5.5 (high effort)

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