A photon and an electron (mass $m$ ) have the same energy $E$. The ratio ( $\lambda_{\text {photon }} / \lambda_{\text {electron }}$ ) of their de Broglie wavelengths is: ( $c$ is the speed of light)
An electron and an alpha particle are accelerated by the same potential difference. Let $$\lambda_e$$ and $$\lambda_\alpha$$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:
If $$\phi$$ is the work function of photosensitive material in $$\mathrm{eV}$$ and light of wavelength of numerical value $$\lambda=\frac{h c}{e}$$ metre, is incident on it with energy above its threshold value at an instant then the maximum kinetic energy of the photo-electron ejected by it at that instant (Take $$h$$-Plank's constant, $$c$$-velocity of light in free space) is (in SI units):
The graph which shows the variation of $$\left(\frac{1}{\lambda^2}\right)$$ and its kinetic energy, $$E$$ is (where $$\lambda$$ is de Broglie wavelength of a free particle):
