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CUET · PHYSICS · PYQ PAPER 2023

The deBroglie wavelength of electron when accelerates through potential \(V\) is given by \(\lambda=\frac{h}{\sqrt{x} \sqrt{V}}\) then x is equal to :
where \(m =\) mass of electron
\(k =\) kinetic energy of electron
\(p =\) momentum
e = charge

  1. A \(2\) p
  2. B 2 me
  3. C 1.2 me
  4. D 2 mk
Verified Solution

Answer & Solution

Correct Answer

(B) 2 me

Step-by-step Solution

Detailed explanation

\(p = \sqrt{2mK}\) \(K = eV\) \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meV}}\) Comparing \(\lambda=\frac{h}{\sqrt{2meV}}\) with \(\lambda=\frac{h}{\sqrt{x} \sqrt{V}}\) \(x = 2me\)
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