PhysicsMedium145×since 2002Q6579An electron accelerated through a potential difference V1V_{1}V1 has a de-Broglie wavelength of λ\lambdaλ. When the potential is changed to V2V_{2}V2, its de-Broglie wavelength increases by 50%50 \%50%. The value of (V1V2)\left(\frac{V_{1}}{V_{2}}\right)(V2V1) is equal toA32\frac{3}{2}23B4C3D94\frac{9}{4}49Check answerSkip