የኬሚስትሪ መግቢያ ፈተና፣ 2017 ዓ.ም. (2025)

The planetary model, with electrons circling a central nucleus the way planets orbit the sun, came from Rutherford and was refined by Bohr. Thomson's picture had no nucleus at all: he described a ball of positive charge with electrons stuck through it, which is the plum pudding model. Pairing Thomson with the planetary model is therefore the mismatch. Option B names Thomson too, but plum pudding is exactly what he proposed, so that row is matched correctly.

Cathode rays are beams of electrons, and a charged particle placed in an electric field is pushed to one side. This particle is not deflected, so it carries no charge. It still has real mass, and the only subatomic particle that is heavy and neutral is the neutron. It adds to the mass number without changing the atomic number. Option A is tempting because protons are heavy as well, but a proton is positive and the field would deflect it.

The atomic number counts the protons, so this atom has 11 protons. The mass number counts protons plus neutrons together, so the neutrons are \( 23 - 11 = 12 \). The nucleus holds 11 protons and 12 neutrons. Option D is tempting if you read 23 as the number of neutrons, but 23 is the total of both kinds of particle, not the neutrons alone.

In the photoelectric effect a single photon must carry at least the work function energy before an electron can leave the metal. Photon energy is \( E = h\nu \), so it is set by frequency, not by how bright the beam is. Blue light has the higher frequency, so its photons carry enough energy while red photons do not. Option D is tempting because a brighter beam does eject more electrons, but that only happens once the frequency is already above the threshold. Below it, no amount of intensity helps.

Use \( h\nu = h\nu_0 + KE_{max} \). The incoming photon carries \( E = 6.63\times10^{-34}\times3.0\times10^{15} = 1.99\times10^{-18} \) J. Take away the kinetic energy: \( h\nu_0 = 1.99\times10^{-18} - 6.63\times10^{-19} = 1.33\times10^{-18} \) J. Then \( \nu_0 = 1.33\times10^{-18} \div 6.63\times10^{-34} = 2.0\times10^{15} \) Hz. Option B carries the right digits but a negative power of ten, and \( 2.0\times10^{-15} \) Hz is not a frequency any light source produces.

Heisenberg's principle says the position and the momentum of a particle cannot both be pinned down at the same moment. The more tightly you fix where the electron is, the more its momentum is left uncertain, because \( \Delta x \cdot \Delta p \ge h/4\pi \). Option A states the opposite. Option D is tempting, but the limit applies to every object. It only matters for very light particles such as electrons, where the uncertainty is large compared with the values themselves.