Question

What are matter waves? Write the de-Broglie formula for matter waves. The de-Broglie wavelength for a photon and an electron are same. Which will have greater energy?

Hint:

Same \(\lambda\Rightarrow\) same momentum \(p=h/\lambda\). Compare \(E_{photon}=pc\) with \(E_{electron}=p^{2}/2m_e\).

Answer 1

Step 1: Matter waves.
Every moving material particle (electron, proton, atom, etc.) has a wave associated with it. These are called matter waves or de Broglie waves. They are not electromagnetic; their wavelength depends on the particle's momentum.

Step 2: de Broglie formula.
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \] where \(h\) is Planck's constant and \(p=mv\) is the momentum.

Step 3: Equal wavelength means equal momentum.
Given that the photon and the electron have the same \(\lambda\). From \(\lambda = h/p\), equal \(\lambda\) implies equal momentum \(p\).

Step 4: Energy of the photon.
For a photon \(E_{ph} = pc = \dfrac{hc}{\lambda}\).

Step 5: Energy of the electron.
For the electron (non-relativistic kinetic energy) \(E_{e} = \dfrac{p^{2}}{2m_e} = \dfrac{h^{2}}{2m_e\lambda^{2}}\).

Step 6: Compare.
\[ \frac{E_{ph}}{E_{e}} = \frac{pc}{p^{2}/2m_e} = \frac{2m_e c}{p} = \frac{2m_e c\,\lambda}{h} \] For any ordinary wavelength this ratio is far greater than 1 (because \(c\) is very large). Hence the photon has the greater energy. \[\boxed{E_{ph} = pc \;\gg\; E_{e}=\frac{p^{2}}{2m_e}\ \Rightarrow\ \text{photon has greater energy}}\]