MHT CET · Physics · Dual Nature of Matter
The kinetic energy of an electron is tripled, then the de-Broglie wavelength associated with it, will change by a factor
- A \(\frac{1}{3}\)
- B \(3\)
- C \(\sqrt{3}\)
- D \(\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
Relation between kinetic energy and momentum is as follows:
\(P=\frac{1}{\sqrt{2 m K}}\)
If kinetic energy is tripled, the new momentum is given by,
\(P^{\prime}=\frac{1}{\sqrt{2 m(3 K)}}=\frac{P}{\sqrt{3}}\)
de-Broglie wavelength of an electron is the following ratio: \(\lambda=\frac{h}{P}\)
Therefore, on tripling the kinetic energy the wavelength will change to: \(\lambda^{\prime}=\frac{h}{P^{\prime}}=\sqrt{3}\left(\frac{h}{P}\right)=\sqrt{3} \lambda\)
\(P=\frac{1}{\sqrt{2 m K}}\)
If kinetic energy is tripled, the new momentum is given by,
\(P^{\prime}=\frac{1}{\sqrt{2 m(3 K)}}=\frac{P}{\sqrt{3}}\)
de-Broglie wavelength of an electron is the following ratio: \(\lambda=\frac{h}{P}\)
Therefore, on tripling the kinetic energy the wavelength will change to: \(\lambda^{\prime}=\frac{h}{P^{\prime}}=\sqrt{3}\left(\frac{h}{P}\right)=\sqrt{3} \lambda\)
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