Question:medium

The de Broglie wavelength for an electron accelerated through the potential difference of \(V_1\) volt is \( \lambda_1 \). When the potential difference is changed to \(V_2\) volt, the associated de Broglie wavelength is increased by \(50%\). If \( \left(\frac{V_1}{V_2}\right)=\frac{9}{\alpha} \), then the value of \( \alpha \) is _____.

Updated On: Apr 12, 2026
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Correct Answer: 4

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