An electron accelerated through a potential difference '$$V_1$$' has a de-Broglie wavelength '$$\lambda$$'. When the potential is changed to '$$V_2$$' its de-Broglie wavelength increases by $$50 \%$$. The value of $$\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)$$ is
- A$$3: 1$$
- B$$9: 4$$
- C$$3: 2$$
- D$$4: 1$$
✓ Correct answer: B
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