የ12ኛ ክፍል ኬሚስትሪ መግቢያ ፈተና፣ 2017 ዓ.ም. (2025)፣ እትም 2

In the cathode ray tube the rays cast a sharp shadow of any object placed in the beam, which shows they travel in straight lines. When Thomson put an electric field and then a magnetic field across the tube, the beam bent, and the direction of the bend showed the particles carry negative charge. Those two observations together are what option D states. Option A is tempting because it names the electron correctly, but it calls the particles positive, and the deflection Thomson measured showed the opposite.

Dalton's biggest error was treating the atom as a solid ball that cannot be divided. To improve on it you need an experiment that looks inside the atom, and the cathode ray tube did exactly that: it produced electrons, particles far smaller than any atom, from whatever metal was used. Option D is the tempting one because isotopes really did refine Dalton's picture of atomic mass, but atoms of one element differ only slightly in mass, not completely, and that experiment leaves the indivisible atom untouched.

Let $x$ be the fraction of A-90, so $1-x$ is the fraction of A-94. The weighted average gives $90x + 94(1-x) = 91$, which becomes $94 - 4x = 91$, so $x = 0.75$. That is 75% A-90 and 25% A-94. A quick sense check: 91 sits much closer to 90 than to 94, so the light isotope has to be the common one. Options A and C put most of the sample on A-94, which would pull the average above 92.

Line spectra exist because an atom's electrons can only sit at fixed energy levels. When an electron takes in energy it jumps to a higher level, and when it drops back it gives out a photon whose energy equals the gap it fell through. Fixed gaps give fixed photon energies, which is why the spectrum is a set of sharp lines instead of a smooth band. Option B is tempting because the Bohr picture has electrons moving around the nucleus, but an electron in a stable level does not radiate; energy is only exchanged when it changes level.

Chromium is $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1$, and the question asks for its electron in an s orbital, which is the 4s one. A 4s orbital means $n = 4$ and $l = 0$, and when $l = 0$ the only allowed value is $m_l = 0$. Set D matches. Option A is tempting because it also has $l = 0$, but $n = 3$ describes a 3s electron buried in an inner shell, not the outermost one.

The prefix M means mega, or $10^6$, so 100 MHz is $1.00 \times 10^8$ Hz. Wavelength comes from $\lambda = c/f$, which gives $\lambda = (3.00 \times 10^8)/(1.00 \times 10^8) = 3.0$ m. Option C is the trap: it comes from dividing by 100 instead of by $10^8$, which means forgetting to convert megahertz into hertz.