If \( \lambda \) and \( K \) are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:
If \( \lambda \) and \( K \) are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:
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The Correct Option is A
Solution and Explanation
To determine the correct graphical representation of the relationship between de Broglie wavelength \( \lambda \) and kinetic energy \( K \) for a particle of constant mass, we start with the de Broglie wavelength formula: \(\lambda = \frac{h}{p}\), where \( h \) is Planck's constant and \( p \) is momentum. The momentum \( p \) of a particle is given by \(\sqrt{2mK}\) for a particle with mass \( m \) and kinetic energy \( K \). Substituting gives:
\[\lambda = \frac{h}{\sqrt{2mK}}\]
This shows an inverse relationship between \(\lambda\) and \(\sqrt{K}\). Squaring both sides results in:
\[\lambda^2 \propto \frac{1}{K}\]
Graphing \(\lambda^2\) vs \(K\) will yield a hyperbolic curve.
The correct representation is 
which accurately depicts \(\lambda^2\) decreasing as \(K\) increases, in line with the inverse proportionality.
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