JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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 \((V_1/V_2) = (9/\alpha)\), then the value of \(\alpha\) is __________.
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
The de Broglie wavelength of an electron accelerated through a potential difference \(V\) is given by \(\lambda = \dfrac{h}{\sqrt{2meV}}\). This implies \(\lambda \propto \dfrac{1}{\sqrt{V}}\). Given that the new wavelength \(\lambda_2\) is increased by \(50\%\), we have…
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