NEET Chemistry Questions: Structure of Atom

Pratcice NEET questions from all capters from huge question bank for free. All MCQs are based on NCERT syllabus. To practice from a specific subject and chapter, select a subject below. Login to practice in a structured way with explanations, bookmarks, lists, notes etc. Click here to Login or Sign up for free.

Physics Chemistry Botany Zoology

Question bank:

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at




$ for Balmer series n_1 = 2 and n_2 = 3 for first line $
The de-Broglie wavelength of a particle with mass 1g and velocity 100 m/s is




No explanation available.
Which of the following sets of quantum numbers belongs to highest energy ?




The principal quantum number (n) determines the energy level, and higher values of n correspond to higher energy levels. The orbital angular momentum quantum number (l) can have values from 0 to n-1. The magnetic quantum number (m) can have values from -l to +l. The spin quantum number (s) can have values of +1/2 or -1/2.
If wavelength of photon is $ 2·2 \time 10^{–11} m, h = 6·6 \times10 ^{–34} Js$ , then momentum of photon is




No explanation available.
According to Bohr’s theory, the energy required for the transition of H atom from n = 6 to n = 8 state is




According to Bohr's theory, the energy of an electron in a hydrogen atom is given by E = -13.6 eV/n^2, where n is the principal quantum number. The energy difference between n=6 and n=8 is less than the energy difference between n=5 and n=7, as the energy levels get closer together for higher values of n.
An electron has kinetic energy of $ 2·14 \times 10^{–22} J.Its de-Broglie wavelength will be nearly (m_e = 9.1 \times 10^{–31} kg) $




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.
What will be de-Broglie wavelength of an electron moving with a velocity of $ 1·20 \times 10^5ms^{–1} $ ?




No explanation available.
The de-Broglie wavelength associated with ball of mass 200 g and moving at a speed of 5 m hour–1 is of the order of $(h = 6·625 \times 10^{–34} Js) $




No explanation available.
The third line of the Balmer series. in the emission spectrum of the hydrogen atom, is due to the transition from the




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).
The highest number of unpaired electrons are w present in




No explanation available.
Rutherford’s atomic model suggests the existence




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.
Which is not true with respect to cathode rays?




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.
In hydrogen atom, energy of first excited state is -3·4 eV. Find out the K.E. of the same orbit of hydrogen atom




K.E. of e- in nth orbit = - $E_n$ = 3.4 e.V
The energy of the first electron in helium will be




$ E_n = - { 13.6 \over n^2 } Z^2 e.V $
In the Bohr’s orbit, what is the ratio of total kinetic energy and total energy of the electron




$ 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 $