A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is :

If Electric field intensity of a uniform plane electromagnetic wave is given as $$E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}{V \over m}$$. Then magnetic intensity 'H' of this wave in Am^$$-$$1 will be :

[Given : Speed of light in vacuum $$c = 3 \times {10^8}$$ ms^$$-$$1, Permeability of vacuum $${\mu _0} = 4\pi \times {10^{ - 7}}$$ NA^$$-$$2]

In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size $${\lambda \over {100}}$$, where $$\lambda$$ is the wavelength of the wave in free space. The phenomenon, which happens there will be :

An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are E_e and p_e and that of photon are E_ph and p_ph respectively. Which of the following is correct?