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
- A \(2\) p
- B 2 me
- C 1.2 me
- D 2 mk
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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