JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
An electron accelerated through a potential difference \(V_1\) has a de-Broglie wavelength of \(\lambda\). When the potential is changed to \(V _2\), its de-Broglie wavelength increases by \(50 \%\). The value of \(\left(\frac{ V _1}{ V _2}\right)\) is equal to :
- A \(3\)
- B \(\frac{9}{4}\)
- C \(\frac{3}{2}\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(\frac{9}{4}\)
Step-by-step Solution
Detailed explanation
\(KE =\frac{ P ^2}{2 m }, \quad P =\frac{ h }{\lambda}\) \(eV _1=\frac{\left(\frac{ h }{\lambda}\right)^2}{2 m }\) \(eV _2=\frac{\left(\frac{ h }{1.5 \lambda}\right)^2}{2 m }\) \(\frac{ V _1}{ V _2}=(1.5)^2=\frac{9}{4}\)
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