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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2022
  • Easy
  • MCQ

The equation \(\lambda=\frac{1.227}{ x } nm\) can be used to find the de-Brogli wavelength of an electron. In this equation \(x\) stands for : Where,\(m =\) mass of electron \(P =\) momentum of electron \(K =\) Kinetic energy of electron \(V =\) Accelerating potential in volts for electron

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Correct Answer

(D) \(\sqrt{ V }\)

Step-by-step Solution

Detailed explanation

\(\lambda=\frac{ h }{ m v}\) (de-Broglie's wavelength)

\(\lambda \frac{ h }{\sqrt{2 m ( K \cdot E )}}\)

\(h =\frac{ h }{\sqrt{2 mqV }}\)

Putting the values of \(m ; q\)

We get \(\lambda=\frac{1 \cdot 22}{\sqrt{V}}\,nm\)

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