Show that the De-Broglie wavelength for a particle with rest mass \(m_0\) and kinetic energy \(K\) is given by: \[ \lambda = \frac{hc}{\sqrt{K(K + 2m_0c^2)}} \]
\[ \lambda = \frac{h}{p} \]
where :
- λ is the de Broglie wavelength.
- h is the Planck constant (\(6.626 \times 10^{-34}\) Js).
- p is the momentum of the particle.
In this derivation, we will calculate the de Broglie wavelength for a particle with rest mass \(m_0\) and kinetic energy \(K\).
DERIVATION
Step 1: Find the momentum of the particle using the relativistic momentum formula :
\[ p = \frac{m_0 \cdot v}{\sqrt{1 - \frac{v^2}{c^2}}} \]
where :
- \( v \) is the velocity of the particle.
- \( c \) is the speed of light in a vacuum (\(3 \times 10^8\) m/s).
Step 2: Find the velocity (\( v \)) of the particle using the kinetic energy formula :
\[ K = \frac{1}{2} m_0 v^2 \]
Step 3: Substitute the expression for velocity (\( v \)) from Step 2 into the momentum formula from Step 1 :
\[ p = \frac{m_0 \cdot \sqrt{2K}}{\sqrt{1 - \frac{2K}{m_0 c^2}}} \]
Step 4: Now, apply the de Broglie wavelength equation to the momentum (\( p \)) obtained in Step 3 :
\[ \lambda = \frac{h}{p} = \frac{h}{\frac{m_0 \cdot \sqrt{2K}}{\sqrt{1 - \frac{2K}{m_0 c^2}}}} \]
Step 5: Simplify the expression in the denominator :
\[ \lambda = \frac{h \sqrt{1 - \frac{2K}{m_0 c^2}}}{m_0 \cdot \sqrt{2K}} \]
Step 6: Rationalize the denominator :
\[ \lambda = \frac{h \sqrt{1 - \frac{2K}{m_0 c^2}}}{m_0 \cdot \sqrt{2K}} \times \frac{\sqrt{1 + \frac{2K}{m_0 c^2}}}{\sqrt{1 + \frac{2K}{m_0 c^2}}} \]
\[ \lambda = \frac{h \sqrt{(1 - \frac{2K}{m_0 c^2})(1 + \frac{2K}{m_0 c^2})}}{m_0 \cdot \sqrt{2K} \cdot \sqrt{1 + \frac{2K}{m_0 c^2}}} \]
Step 7: Simplify the expression in the numerator :
\[ \lambda = \frac{h \sqrt{1 - \frac{4K^2}{(m_0 c^2)^2}}}{m_0 \cdot \sqrt{2K} \cdot \sqrt{1 + \frac{2K}{m_0 c^2}}} \]
Step 8: Finally, cancel out \( K \) from the numerator and denominator :
\[ \lambda = \frac{hc}{m_0 c^2} \cdot \frac{1}{\sqrt{1 + \frac{2K}{m_0 c^2}}} \]
\[ \lambda = \frac{hc}{\sqrt{K(K + 2m_0c^2)}} \]
SUMMARY
In this derivation, we have shown that the de Broglie wavelength (\( \lambda \)) for a particle with rest mass \( m_0 \) and kinetic energy \( K \) is given by :
\[ \lambda = \frac{hc}{\sqrt{K(K + 2m_0c^2)}} \]
This result demonstrates the wave-like nature of particles and provides an important insight into the behavior of matter at quantum scales.
Also Read : Explain the Joule-Thompson Effect and Derive the Joule-Thompson Coefficient for a real gas.
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