Question:
If 2E is the kinetic energy of a moving particle of mass \(m\), then the wavelength of the de Broglie wave associated with it is
If 2E is the kinetic energy of a moving particle of mass \(m\), then the wavelength of the de Broglie wave associated with it is
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Always express momentum in terms of kinetic energy: \(p = \sqrt{2mK}\) before substituting into \(\lambda = h/p\).
Updated On: Apr 24, 2026
- \(\frac{h}{2mE}\)
- \(\frac{h}{mE}\)
- \(\frac{h}{2\sqrt{mE}}\)
- \(\frac{h}{\sqrt{2mE}}\)
- \(\frac{h}{4\sqrt{mE}}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
de Broglie wavelength: \(\lambda = \frac{h}{p}\), where \(p\) is momentum. Kinetic energy \(K = \frac{p^2}{2m}\).
Step 2: Detailed Explanation:
Given \(K = 2E\). So, \(2E = \frac{p^2}{2m} \Rightarrow p^2 = 4mE \Rightarrow p = 2\sqrt{mE}\).
Thus, \(\lambda = \frac{h}{p} = \frac{h}{2\sqrt{mE}}\).
Step 3: Final Answer:
The de Broglie wavelength is \(\frac{h}{2\sqrt{mE}}\).
de Broglie wavelength: \(\lambda = \frac{h}{p}\), where \(p\) is momentum. Kinetic energy \(K = \frac{p^2}{2m}\).
Step 2: Detailed Explanation:
Given \(K = 2E\). So, \(2E = \frac{p^2}{2m} \Rightarrow p^2 = 4mE \Rightarrow p = 2\sqrt{mE}\).
Thus, \(\lambda = \frac{h}{p} = \frac{h}{2\sqrt{mE}}\).
Step 3: Final Answer:
The de Broglie wavelength is \(\frac{h}{2\sqrt{mE}}\).
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