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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2023
  • Medium
  • MCQ

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 :

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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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