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Module 5 : MODERN PHYSICS

Lecture 26 : Wave Nature of Particle - the de Broglie Hypothesis

	Thus
	$\displaystyle p_x^\prime-\frac{h}{\lambda^\prime}\sin\theta =p_x^{\prime\prime}+ \frac{h}{\lambda^{\prime\prime}}\sin\theta$
	so that the momemntum uncertainty of the electron is
	$\displaystyle \Delta p_x = p_x^\prime-p_x^{\prime\prime}=\frac{2h}{\lambda}\sin\theta=\frac{h}{\Delta x}$
	giving
	$\displaystyle \Delta x\Delta p_x \approx h$
	It may be noted that one can always determine the momentum along the y-direction with any desired degree of accuracy when there is position uncertainty along the x-direction.
	Exercise 8
	Example 17
	Example 18