The de-Broglie wavelength of a particle with mass 1g and velocity 100 m/s is
Atomic Structure MCQs for NEET — Chemistry Questions with Answers
Practice free Atomic Structure (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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
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} $ ?
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) $
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
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.
Ready to ace NEET?
Free access · No credit card required
Frequently Asked Questions
Yes. You can attempt every Atomic Structure question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.