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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter

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

  1. A \(\sqrt{ mK }\)
  2. B \(\sqrt{ P }\)
  3. C \(\sqrt{ K }\)
  4. D \(\sqrt{ V }\)
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Correct Answer

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

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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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