Question

de Broglie wavelength \( \lambda \) as a function of \( \frac{1}{\sqrt{K}} \), for two particles of masses \( m_1 \) and \( m_2 \), is shown in the figure. Here, \( K \) is the energy of the moving particles. (a) What does the slope of a line represent? 
(b) Which of the two particles is heavier? 
(c) Is this graph also valid for a photon? Justify your answer in each case.

 

Hint:

For massive particles, the de Broglie wavelength follows \( \lambda = \frac{h}{\sqrt{2mK}} \), while for photons, momentum is given by \( p = \frac{h}{\lambda} \), making the given relation inapplicable.

Answer 1

(a) The de Broglie wavelength is given by: \[ \lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2m}} \times \frac{1}{\sqrt{K}} \] Comparing with the equation of a straight line (\( y = mx + c \)), the slope of the line is: \[ \text{slope} = \frac{h}{\sqrt{2m}} \] Thus, the slope of the line represents \( \frac{h}{\sqrt{2m}} \), which is inversely proportional to \( \sqrt{m} \). (b) From the equation: \[ \text{slope} \propto \frac{1}{\sqrt{m}} \] Since the slope of \( m_2 \) is greater than that of \( m_1 \), it implies that \( m_2 \) is lighter, and \( m_1 \) is heavier. (c) No, this graph is not valid for a photon. The given equation for momentum: \[ p = \sqrt{2mK} \] is not applicable to a photon, as a photon has zero rest mass and its momentum is given by: \[ p = \frac{h}{\lambda} \]