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 Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?
How many alpha and beta particles are emitted when Uranium 92U238 decays to lead 82Pb206 ?