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 expressed as \[\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2m}} \times \frac{1}{\sqrt{K}}\]. When compared to a linear equation \( y = mx + c \), the slope is determined to be \[\text{slope} = \frac{h}{\sqrt{2m}}\]. Therefore, the slope signifies \( \frac{h}{\sqrt{2m}} \) and exhibits an inverse proportionality to \( \sqrt{m} \).

(b) From the equation \[\text{slope} \propto \frac{1}{\sqrt{m}}\], a steeper slope for \( m_2 \) compared to \( m_1 \) indicates that \( m_2 \) is less massive (lighter) and \( m_1 \) is more massive (heavier).

(c) This graph is not applicable to a photon. The equation for momentum, \( p = \sqrt{2mK} \), does not apply to a photon, which has zero rest mass. The momentum of a photon is defined as \( p = \frac{h}{\lambda} \).