Chemistry MCQs for NEET — Practice Questions with Answers

The de Broglie wavelength of a particle is given by λ = h / (mv), where h is Planck's constant, m is the mass of the particle, and v is its velocity. For an electron with kinetic energy of 2.14 × 10^-22 J, we can calculate its velocity using the relation KE = (1/2)mv^2. Substituting the values, we get λ = 9.28 × 10^-8 m.

The Balmer series in the hydrogen spectrum corresponds to the electron transitions from higher energy levels to the second energy level (n=2). The third line in the Balmer series is due to the transition from the fifth orbit (n=5) to the second orbit (n=2).

Rutherford's alpha particle scattering experiment led to the discovery of the atomic nucleus. His model proposed that the positive charge and most of the mass of an atom is concentrated in a tiny region called the nucleus, surrounded by empty space with orbiting electrons.

Cathode rays are a stream of electrons emitted from the cathode of a vacuum tube. They are charged particles that can be deflected by electric or magnetic fields. However, they do not move with the speed of light, which is the maximum possible speed in the universe.

K.E. of e- in nth orbit = - $E_n$ = 3.4 e.V

$ E_n = - { 13.6 \over n^2 } Z^2 e.V $

$ K.E = { 1 \over 2 } mv ^2 P.E = - { Ze^2 \over r } $ Electrostatic force = centrifugal force $ { Ze^2 \over r^2 } = {mv^2 \over r } \therefore P.E = -mv^2 $ $ Total energy = K.E + P.E = { 1 \over 2 } mv^2 - mv^2 = -{ 1 \over 2 }mv^2 \therefore { K.E \over Total Energy } = -1 $