Question

A photon and electron have got the same de Broglie wavelength. Explain which has greater total energy.

Answer 1

We know that,

For electron

$E=m{c}^2.....\left(I\right)$

Also,

$\lambda =\frac{h}{mv}$

$m=\frac{h}{\lambda v}$

Now, put the value of m in equation (I)

$E=\frac{h}{\lambda v}\times c^2$

$E=\frac{h{c}^2}{\lambda v}$

Now, for photon

$E^{\prime }=h\nu =h\frac{c}{\lambda }...\left(II\right)$

Now, dividing equation (I) by (II)

$\frac{E}{E^{\prime }}=\frac{h{c}^2\lambda }{hc\times \lambda v}$

$\frac{E}{E^{\prime }}=\frac{c}{v}$

Now,

$\frac{E}{E^{\prime }}=\frac{c}{v}$

$c>v\frac{E}{E^{\prime }}>1$

$E>E^{\prime }$

The total energy of an electron is greater than the total energy of a photon

Hence, this is the required solution