If the K.E. of an electron and a photon is the same, then the relation between their de-Broglie wavelength will be:

1. ${\lambda }_{ph}$ $<$ ${\lambda }_e$

2. ${\lambda }_{ph}$ $=$ ${\lambda }_e$

3. ${\lambda }_{ph}$ $>$ ${\lambda }_e$

4. ${\lambda }_{ph}$ $=2$ ${\lambda }_e$

If particles are moving with the same velocity, then the de-Broglie wavelength is maximum for:
1. proton
2. \(\alpha-\)particle
3. neutron
4. \(\beta-\)particle