An $$\alpha$$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $$\lambda$$. To double the kinetic energy, the incident light must have wavelength:

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

The equation $$\lambda=\frac{1.227}{x} \mathrm{~nm}$$ can be used to find the de-Brogli wavelength of an electron. In this equation $$x$$ stands for :

Where

$$\mathrm{m}=$$ mass of electron

$$\mathrm{P}=$$ momentum of electron

$$\mathrm{K}=$$ Kinetic energy of electron

$$\mathrm{V}=$$ Accelerating potential in volts for electron