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If the linear momentum of a particle is 2.2×10⁴ kg m s⁻¹, what will be its de Broglie wavelength? (Take h = 6.6×10⁻³⁴ J s)

Asked in GUJCET 2009 · de Broglie relation and its graphs

• (1) 3×10⁻²⁹ nm
• (2) 6×10⁻²⁹ nm
• (3) 3×10⁻²⁹ m
• (4) 6×10⁻²⁹ m

Step-by-step solution

Why the other options are wrong

• (2) No single slip gives twice the correct value. Sanity check: (6.6)/(2.2) = 3, so λ = 3×10⁻³⁸ m = 3×10⁻²⁹ nm.
• (3) Right number, wrong unit. The wavelength is 3×10⁻³⁸ m, which is 3×10⁻²⁹ nm; 3×10⁻²⁹ m is a billion times too long.
• (4) No single slip gives this: it is twice the correct number and in the wrong unit. Sanity check: λ = (6.6×10⁻³⁴)/(2.2×10⁴) = 3×10⁻³⁸ m.

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