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Qormaata galuumsa kutaa 12ffaa Keemistirii, bara 2017 A.L.I (2025), maxxansa 2

Gaaffilee dhugaa qormaata galuumsa yunivarsiitii kutaa 12 kan Keemistirii, akkuma barattoonni bara 2017 A.L.I itti fudhatanitti. Tokkoon tokkoon gaaffii deebii sirrii fi ibsa qaba. Qormaanni bara 2017 A.L.I maxxansa adda addaatiin bahe. Fuulli kun maxxansa 2 qaba; kaan gadi aanee tarreeffamaniiru.

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Gaaffii 1

Which of the following is correct about the postulates of Thomson's experiment?

  1. A.Cathode rays are made of positively charged particles, later called electrons.
  2. B.Electrons are NOT present in all atoms of all elements.
  3. C.Electrons have a very large mass compared to the atom.
  4. D.Cathode rays travel in straight lines and can be deflected by electric and magnetic fields
Deebii fi ibsa ilaalaa

Deebii: D Cathode rays travel in straight lines and can be deflected by electric and magnetic fields

Maaliif

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.

Gaaffii 2

Imagine you are a scientist in the 19th century and want to improve Dalton's Atomic Theory. Which of the following new experiments would best refine his ideas?

  1. A.Designing a cathode ray tube experiment to identify subatomic particles.
  2. B.Proposing that all atoms are made of fire, water, air, and earth
  3. C.Conducting an experiment to prove that atoms cannot form compounds.
  4. D.Showing that atoms of the same element have completely different masses.
Deebii fi ibsa ilaalaa

Deebii: A Designing a cathode ray tube experiment to identify subatomic particles.

Maaliif

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.

Gaaffii 3

The atomic mass of element A is 91 atomic mass unit. The only naturally occurring isotopes of element A are A-90 and A-94. The percent abundance in naturally occurring sample of element A are

  1. A.31% A-90 and 69% A-94
  2. B.75% A-90 and 25% A-94
  3. C.69% A-94 and 31% A-90
  4. D.20% A-94 and 80% A-90
Deebii fi ibsa ilaalaa

Deebii: B 75% A-90 and 25% A-94

Maaliif

Let xx be the fraction of A-90, so 1−x1-x is the fraction of A-94. The weighted average gives 90x+94(1−x)=9190x + 94(1-x) = 91, which becomes 94−4x=9194 - 4x = 91, so x=0.75x = 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.

Gaaffii 4

Which of the following statements best explains the relationship between electrons' energy levels and the emission spectra of atoms?

  1. A.Electrons absorb energy on jumping from lower to higher energy level
  2. B.Electron emit energy on revolving around the nucleus
  3. C.Electron absorb energy on revolving around the nucleus
  4. D.The emission spectra are unrelated to electron transitions but depend on atomic size.
Deebii fi ibsa ilaalaa

Deebii: A Electrons absorb energy on jumping from lower to higher energy level

Maaliif

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.

Gaaffii 5

Which one of the following sets of quantum numbers could be those of the distinguishing (last) electron of Cr at s-orbital?

  1. A.n = 3, l = 0, m_l = 0, m_s = +1/2
  2. B.n = 3, l = 2, m_l = -1, m_s = +1/2
  3. C.n = 4, l = 1, m_l = 3, m_s = -1/2
  4. D.n = 4, l = 0, m_l = 0, m_s = -1/2
Deebii fi ibsa ilaalaa

Deebii: D n = 4, l = 0, m_l = 0, m_s = -1/2

Maaliif

Chromium is 1s2 2s2 2p6 3s2 3p6 3d5 4s11s^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=4n = 4 and l=0l = 0, and when l=0l = 0 the only allowed value is ml=0m_l = 0. Set D matches. Option A is tempting because it also has l=0l = 0, but n=3n = 3 describes a 3s electron buried in an inner shell, not the outermost one.

Gaaffii 6

A radio station broadcasts at a frequency of 100 MHz. What is the wavelength of the radio waves? (Speed of light, c = 3.00 × 10⁸ m/s)

  1. A.3.0 × 10⁻³ m
  2. B.3.0 m
  3. C.3.0 × 10⁶ m
  4. D.3.0 × 10⁻³ m
Deebii fi ibsa ilaalaa

Deebii: B 3.0 m

Maaliif

The prefix M means mega, or 10610^6, so 100 MHz is 1.00×1081.00 \times 10^8 Hz. Wavelength comes from λ=c/f\lambda = c/f, which gives λ=(3.00×108)/(1.00×108)=3.0\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 10810^8, which means forgetting to convert megahertz into hertz.

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Kunneen gaaffilee dhugaa qormaata galuumsa Keemistirii bara 2017 A.L.I ti?

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