The de Broglie wavelength λ associated with an electron accelerated through a potential difference of 121 V is . (Take mₑ = 9.1×10⁻³¹ kg and h = 6.63×10⁻³⁴ J s.)

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(3) 1.12 A. Given: V = 121 V, mₑ = 9.1×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C, h = 6.63×10⁻³⁴ J s. Idea: the electron gains kinetic energy eV, so its momentum is p = √2mₑ eV and λ = h/(√2mₑ eV). 2mₑ eV = 2×9.1×10⁻³¹×1.6×10⁻¹⁹×121 = 3.52×10⁻⁴⁷, so p = 5.94×10⁻²⁴ kg m s⁻¹. λ = (6.63×10⁻³⁴)/(5.94×10⁻²⁴) = 1.12×10⁻¹⁰ m Shortcut: λ = (12.27)/(√V) A = (12.27)/(11) = 1.12 A. So λ ≈ 1.12 A.

The de Broglie wavelength λ associated with an electron accelerated through a potential difference of 121 V is ______. (Take mₑ = 9.1×10⁻³¹ kg and h = 6.63×10⁻³⁴ J s.)

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