The $$\beta$$-decay process, discovered in around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e^$$-$$) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, that is, n $$\to$$ p + e^$$-$$ + $${\overline v _e}$$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($${\overline v _e}$$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 $$\times$$ 10⁶ eV. The kinetic energy carried by the proton is only the recoil energy.

If the anti-neutrino had a mass of 3 eV/c² (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?

Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, $${{\sin {\theta _1}} \over {\sin {\theta _2}}} = {{{n_2}} \over {{n_1}}}$$, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation $$n = \left( {{c \over v}} \right) = \pm \,\sqrt {{\varepsilon _r}{\mu _r}} $$, where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, $$\varepsilon $$_r and $$\mu$$_r, are the relative permittivity and permeability of the medium, respectively.

In normal materials, both $$\varepsilon$$_r and $$\mu$$_r , are positive, implying positive n for the medium. When both $$\varepsilon$$_r and $$\mu$$_r are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating and physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

For light incident from air on a meta-material, the appropriate ray diagram is

Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, $${{\sin {\theta _1}} \over {\sin {\theta _2}}} = {{{n_2}} \over {{n_1}}}$$, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation $$n = \left( {{c \over v}} \right) = \pm \,\sqrt {{\varepsilon _r}{\mu _r}} $$, where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, $$\varepsilon $$_r and $$\mu$$_r, are the relative permittivity and permeability of the medium, respectively.

In normal materials, both $$\varepsilon$$_r and $$\mu$$_r , are positive, implying positive n for the medium. When both $$\varepsilon$$_r and $$\mu$$_r are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating and physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

Choose the correct statement.

In the given circuit, the AC source has $$\omega$$ = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)