Questions: Structure of Atom

Q1 What did Rutherford's alpha particle scattering experiment show, and how did it contradict the Thomson model?

Answer (English): Alpha particles were fired at a thin gold foil. Most passed straight through, a few were deflected through small angles, and a very small fraction came back. Under the Thomson model the positive charge is spread over the whole atom, so a fast alpha particle would pass through with only a tiny deflection and backward scattering would never happen. Rutherford concluded that nearly all the mass and the whole positive charge are concentrated in a small central nucleus, with electrons occupying the rest of the atom's volume. His estimate put the nuclear radius near 10^-15 m against an atomic radius near 10^-10 m.

Hinglish: Rutherford ne gold foil par alpha particles daage. Zyada particles seedhe, kuch chhote angle par, aur bahut thode wapas aaye. Thomson model mein charge poore atom mein faila hota, isliye itna bada deflection possible nahi tha. Isse pata chala ki mass aur positive charge chhote nucleus mein concentrated hai; nucleus ka radius ~10^-15 m aur atom ka ~10^-10 m hai.

Q2 State the three main observations of the photoelectric effect and explain them with Einstein's photon picture.

Answer (English): First, emission begins almost immediately when light falls, with no measurable delay. Second, the maximum kinetic energy of the ejected electrons depends on the frequency of the light and not on its intensity. Third, below a threshold frequency no electrons are emitted at all, however bright the light is. Einstein explained all three by treating light as photons of energy h nu. One photon is absorbed by one electron, so the maximum kinetic energy is the photon energy minus the work function. Increasing intensity adds more photons and so more electrons, but each electron still takes only one photon.

Hinglish: Teen observations: emission turant hoti hai; maximum kinetic energy light ki frequency par depend karti hai, intensity par nahi; aur threshold frequency se neeche koi emission nahi hoti. Einstein ne kaha light photons se banti hai, har photon ki energy hν; ek electron ek photon leta hai. Isliye K_max = hν - W0; intensity badhane se sirf electrons ki sankhya badhti hai.

Q3 A photon has an energy of 3.0 eV. Find its wavelength and frequency.

Answer (English): Convert the energy to joules: 3.0 x 1.602 x 10^-19 = 4.81 x 10^-19 J. Wavelength is hc divided by E. Using the convenient value hc = 1.986 x 10^-25 J m gives 1.986 x 10^-25 divided by 4.81 x 10^-19, which is 4.13 x 10^-7 m, or 413 nm. The frequency is c divided by the wavelength, 3.0 x 10^8 divided by 4.13 x 10^-7, which is 7.27 x 10^14 Hz. The light lies in the violet end of the visible spectrum.

Hinglish: Energy ko joule mein badlo: 3.0 eV = 4.81 × 10^-19 J. λ = hc/E se 413 nm milta hai. ν = c/λ se 7.27 × 10^14 Hz aata hai. Yeh violet ke paas visible light hai.

Q4 Calculate the radius and the energy of the third Bohr orbit of hydrogen.

Answer (English): The radius is 52.9 n squared pm divided by Z, and for hydrogen Z is 1, so the third orbit has radius 52.9 x 9 = 476.1 pm. The energy is -13.6 Z squared divided by n squared eV, which gives -13.6/9 = -1.51 eV. The negative sign shows the electron is still bound. The energy is less negative than the ground state, so the third orbit is higher in energy.

Hinglish: H atom ke liye r_n = 52.9 n² pm se r_3 = 476 pm. E_n = -13.6/n² eV se E_3 = -1.51 eV. Negative sign matlab electron bound hai; ground state -13.6 eV se yeh zyada high energy state hai.

Q5 Why does hydrogen give a line spectrum instead of a continuous spectrum? Name the three main series and the region each falls in.

Answer (English): The electron in a hydrogen atom can have only the energies allowed by the Bohr condition, so it can absorb or emit only photons whose energy equals the gap between two levels. A continuous spectrum would need every possible energy gap, which the discrete levels do not provide. The Lyman series ends on n = 1 and lies in the ultraviolet. The Balmer series ends on n = 2 and has four lines in the visible region. The Paschen series ends on n = 3 and lies in the infrared. The Rydberg formula predicts the positions of all these lines.

Hinglish: Hydrogen ke electron ki energy levels fixed hoti hain, isliye sirf unke beech ke gap ke photons absorb ya emit hote hain, aur line spectrum milta hai. Lyman series n=1 par khatam hoti hai aur UV mein, Balmer n=2 par (visible), aur Paschen n=3 par (infrared). Rydberg formula in lines ki position deta hai.

Q6 A cricket ball of mass 0.15 kg moves at 30 m s^-1. Calculate its de Broglie wavelength and explain why wave behaviour is not noticed.

Answer (English): The de Broglie wavelength is h divided by mv. Substituting 6.626 x 10^-34, 0.15 and 30 gives 1.5 x 10^-34 m. This is far smaller than any gap the ball could diffract through, so no wave behaviour is observed. For an electron the mass is about 10^-30 kg and the wavelength can be of the order of 10^-10 m, which is comparable with atomic spacing, so electron diffraction is observable.

Hinglish: λ = h/mv se ball ki wavelength 1.5 × 10^-34 m aati hai, jo bohot chhoti hai, isliye diffraction nahi dikhta. Electron ka mass ~10^-30 kg hota hai, isliye uski wavelength ~10^-10 m tak aati hai aur crystal se diffraction dikhta hai.

Q7 State Heisenberg's uncertainty principle and explain one consequence for the Bohr model.

Answer (English): The product of the uncertainty in position and the uncertainty in momentum of a particle cannot be less than h divided by 4 pi. The relation follows from the wave nature of matter and does not depend on the quality of the measuring instrument. A sharply defined position requires a broad range of wavelengths, which means a wide spread of momentum. The Bohr model assigns a definite circular path with a definite position and speed at every instant, so it conflicts with this principle. The quantum mechanical model replaces the path with an orbital that gives only probabilities.

Hinglish: Uncertainty principle: Δx·Δp ≥ h/4π. Position jitni exact, momentum utna hi uncertain. Bohr model definite path deta hai, jo is principle se conflict karta hai. Modern model path ki jagah orbital deta hai, jo sirf probability batata hai.

Q8 List the four quantum numbers and give all possible values for an electron in a 3p orbital.

Answer (English): The principal quantum number n fixes the shell and the main energy, the azimuthal quantum number l fixes the subshell and shape, the magnetic quantum number m_l fixes orientation, and the spin quantum number m_s fixes the spin direction. For a 3p electron, n is 3, l is 1, m_l can be -1, 0 or +1, and m_s can be +1/2 or -1/2. The three possible m_l values correspond to the three p orbitals, and each orbital can hold two electrons with opposite spins.

Hinglish: Char quantum numbers: n (shell), l (subshell/shape), m_l (orientation), m_s (spin). 3p electron ke liye n = 3, l = 1, m_l = -1, 0, +1, aur m_s = ±1/2. Teen m_l values teen p orbitals dete hain; har orbital mein do electrons opposite spin ke.

Q9 State the Aufbau principle, Pauli exclusion principle and Hund's rule. Why is the configuration of chromium [Ar] 3d5 4s1?

Answer (English): The Aufbau principle fills the lowest energy orbital first. The Pauli exclusion principle forbids two electrons from having the same four quantum numbers, so an orbital holds at most two electrons with opposite spins. Hund's rule says that degenerate orbitals are singly occupied with parallel spins before pairing begins. Chromium has 24 electrons. The simple filling order reaches 3d4 4s2, but moving one electron from 4s to 3d gives a half-filled d subshell. The half-filled arrangement has extra exchange stability and lower energy, so the observed configuration is [Ar] 3d5 4s1.

Hinglish: Aufbau mein lowest energy orbital pehle bharta hai; Pauli kehta hai ek orbital mein do electrons max, opposite spin; Hund kehta hai degenerate orbitals pehle single electrons se bharte hain. Chromium mein 3d4 4s2 ki jagah ek electron 4s se 3d mein jaakar half-filled 3d5 banata hai, jo zyada stable hota hai, isliye configuration [Ar] 3d5 4s1 hai.

Q10 Calculate the energy change and wavelength for the transition from n = 3 to n = 2 in hydrogen.

Answer (English): The energy difference is 13.6 eV multiplied by (1/4 minus 1/9). The bracket equals 5/36, so the energy change is 13.6 x 5/36, which is 1.89 eV. Convert to joules: 1.89 x 1.602 x 10^-19 = 3.03 x 10^-19 J. The wavelength is hc divided by the energy, or 1240 divided by 1.89, which is 656 nm. This is the red line of the Balmer series.

Hinglish: ΔE = 13.6(1/4 - 1/9) = 13.6 × 5/36 = 1.89 eV = 3.03 × 10^-19 J. λ = 1240/E se 656 nm milta hai, jo Balmer series ki red line hai.

That closes the chapter questions. Try the quick quiz below before moving on to the previous year questions.