MCQEasyJEE Main 2023 · 30 January, Shift 2de Broglie Relation

Physics Question from JEE Main 2023 · 30 January, Shift 2

An electron accelerated through a potential difference V1V_1 has a de-Broglie wavelength of λ\lambda. When the potential is changed to V2V_2, its de-Broglie wavelength increases by 50%50\%. The value of V1V2\frac{V_1}{V_2} is equal to :

  • A

    33

  • B

    94\frac{9}{4}

  • C

    32\frac{3}{2}

  • D

    44

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