Questions: Atoms

Q1. Choose the correct alternatives for statements about the Thomson and Rutherford models.

Answer (English): (a) The atom's size is the same in both models, since both describe the same atom. (b) In the ground state of Thomson's model electrons are in stable equilibrium, while in Rutherford's model they always experience a net force. (c) A classical Rutherford atom must collapse, because the orbiting electron radiates energy. (d) The mass distribution is nearly continuous in Thomson's model and highly non-uniform in Rutherford's model. (e) The positive part holds most of the mass in Rutherford's model.

Hinglish: (a) Dono models atom ke size ke baare mein same hain. (b) Thomson model ke ground state mein electrons stable equilibrium mein hain, Rutherford model mein un par hamesha net force lagta hai. (c) Rutherford model par based classical atom collapse ho jaayega kyunki ghoomta electron energy radiate karta hai. (d) Mass distribution Thomson model mein lagbhag continuous hai aur Rutherford model mein bahut non-uniform. (e) Positive charge ke paas zyada mass Rutherford model mein hoti hai.

Q2. What changes in the result if the gold foil is replaced by a thin sheet of solid hydrogen?

Answer (English): Hydrogen nuclei carry only one unit of positive charge, against 79 units for gold, and the mass of a proton is far less than that of a gold nucleus. The repulsive force on an alpha particle would be much weaker, so the distance of closest approach would be larger and the number of alpha particles scattered through large angles would fall sharply. Backward scattering would become very rare, and the evidence for a concentrated nucleus would be much harder to see.

Hinglish: Hydrogen nucleus par sirf ek unit positive charge hota hai, gold par 79, aur proton ka mass gold nucleus se bahut kam hai. Alpha particle par repulsive force bahut kamzor hogi, isliye closest approach ki distance badh jaayegi aur bade angles par scattered particles ki sankhya tezi se gir jaayegi. Backward scattering bahut rare ho jaayega aur concentrated nucleus ka evidence dekhna mushkil hoga.

Q3. Define distance of closest approach and impact parameter, and give the formula for the distance of closest approach.

Answer (English): The impact parameter is the perpendicular distance between the initial velocity line of the alpha particle and the centre of the nucleus. It decides how strongly the particle is deflected. The distance of closest approach is the centre-to-centre separation when a head-on alpha particle momentarily stops and reverses. At that point all its kinetic energy K has become electrical potential energy, so K = (1/4πε0)(2e)(Ze)/d and therefore d = (1/4πε0)(2Ze squared)/K.

Hinglish: Impact parameter alpha particle ke initial velocity ki line aur nucleus ke centre ke beech ki perpendicular distance hai; isse deflection ki strength decide hoti hai. Distance of closest approach head-on collision mein centre-to-centre separation hai jab particle ruk kar wapas mudta hai. Us waqt poori kinetic energy K potential energy ban jaati hai, isliye K = (1/4πε0)(2e)(Ze)/d aur d = (1/4πε0)(2Ze²)/K.

Q4. What are the values of the impact parameter for scattering angles of 0 degrees and 180 degrees?

Answer (English): For a scattering angle of 0 degrees the alpha particle passes far from the nucleus without deviation, so the impact parameter is very large, formally infinite. For 180 degrees the particle undergoes a head-on collision and reverses, so the impact parameter is zero.

Hinglish: 0 degree ke scattering angle par alpha particle nucleus se bahut door se bina deviation nikal jaata hai, isliye impact parameter bahut bada, formally infinite hota hai. 180 degree par head-on collision hota hai aur particle wapas mudta hai, isliye impact parameter zero hota hai.

Q5. Calculate the frequency of the photon when an electron jumps between two levels separated by 2.3 eV.

Answer (English): The photon energy is E = 2.3 eV = 2.3 × 1.6 × 10 to the -19 J = 3.68 × 10 to the -19 J. Since hν = E, the frequency is ν = 3.68 × 10 to the -19/6.6 × 10 to the -34, which is about 5.6 × 10 to the 14 Hz. This falls in the visible range.

Hinglish: Photon energy E = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ J = 3.68 × 10⁻¹⁹ J. hν = E se ν = 3.68 × 10⁻¹⁹/6.6 × 10⁻³⁴ ≈ 5.6 × 10¹⁴ Hz. Yeh visible range mein aati hai.

Q6. For hydrogen, the ground-state energy is E1 = -13.6 eV. Use the relation between total, kinetic and potential energy to find K and U in this state.

Answer (English): For an electron bound in the Coulomb field of a nucleus, the kinetic energy is the negative of the total energy and the potential energy is twice the total: K = -E and U = 2E. With E = -13.6 eV, K = 13.6 eV and U = -27.2 eV. The sum gives -13.6 eV, which agrees with the total energy.

Hinglish: Coulomb field mein bound electron ke liye K = -E aur U = 2E hota hai. E = -13.6 eV rakhne par K = 13.6 eV aur U = -27.2 eV. Dono ka sum -13.6 eV deta hai, jo total energy se match karta hai.

Q7. A hydrogen atom in the ground level absorbs a photon and is excited to n = 4. Find the wavelength and frequency of the photon.

Answer (English): The energy of the nth level is -13.6/n squared eV, so E1 = -13.6 eV and E4 = -0.85 eV. The photon energy is the difference, 12.75 eV. Its wavelength follows from 1240/λ in nm = E in eV, so λ = 1240/12.75 = 97.3 nm. The frequency is hν = E, giving ν = 12.75 × 1.6 × 10 to the -19 divided by 6.6 × 10 to the -34, about 3.1 × 10 to the 15 Hz. The line belongs to the Lyman series because the electron lands on n = 1.

Hinglish: En = -13.6/n² eV se E1 = -13.6 eV aur E4 = -0.85 eV. Photon energy 12.75 eV. Wavelength ke liye 1240/λ(nm) = E(eV), isliye λ = 1240/12.75 = 97.3 nm. Frequency ν = E/h se ≈ 3.1 × 10¹⁵ Hz. Line Lyman series ki hai kyunki electron n = 1 par girta hai.

Q8. Find the speed of the electron in the n = 1, 2 and 3 levels of hydrogen, and the orbital period in each level.

Answer (English): The speed is v n = 2.18 × 10 to the 6/n m/s, so v1 = 2.18 × 10 to the 6 m/s, v2 = 1.09 × 10 to the 6 m/s and v3 = 7.27 × 10 to the 5 m/s. The period is T = 2π r n/v n. With r n = n squared a0 and a0 = 5.3 × 10 to the -11 m, T1 = 2π(5.3 × 10 to the -11)/(2.18 × 10 to the 6), about 1.53 × 10 to the -16 s. Since r n goes as n squared and v n as 1/n, the period goes as n cubed: T2 = 8 T1 and T3 = 27 T1.

Hinglish: Speed vn = 2.18×10⁶/n m/s se v1 = 2.18×10⁶, v2 = 1.09×10⁶ aur v3 = 7.27×10⁵ m/s. Period T = 2πrn/vn; rn = n²a0 aur a0 = 5.3×10⁻¹¹ m rakhne par T1 ≈ 1.53×10⁻¹⁶ s. rn n² ke saath aur vn 1/n ke saath badalta hai, isliye period n³ ke saath: T2 = 8T1 aur T3 = 27T1.

Q9. The innermost orbit of hydrogen has radius 5.3 × 10 to the -11 m. Work out the radii for n = 2 and n = 3.

Answer (English): The radius of the nth orbit is r n = n squared a0. For n = 2, r2 = 4 × 5.3 × 10 to the -11 = 2.1 × 10 to the -10 m. For n = 3, r3 = 9 × 5.3 × 10 to the -11 = 4.8 × 10 to the -10 m. The radius grows as the square of the quantum number.

Hinglish: rn = n²a0 hota hai. n = 2 par r2 = 4 × 5.3×10⁻¹¹ = 2.1×10⁻¹⁰ m, aur n = 3 par r3 = 9 × 5.3×10⁻¹¹ = 4.8×10⁻¹⁰ m. Radius quantum number ke square ke saath badhta hai.

Q10. A 12.5 eV electron beam bombards gaseous hydrogen at room temperature. Which spectral series are emitted?

Answer (English): The ground state is -13.6 eV. An electron with 12.5 eV can raise the atom to a level whose energy is -13.6 + 12.5 = -1.1 eV. Comparing with -13.6/n squared, the energy -1.1 eV is between E3 = -1.51 eV and E4 = -0.85 eV, so the atom can be excited up to n = 3 but not to n = 4. Transitions from n = 2 and n = 3 down to lower levels can then occur, so the Lyman series (to n = 1) and the Balmer series (to n = 2) are emitted. The atom cannot reach n = 4, so no Brackett or higher series lines appear from a single 12.5 eV hit.

Hinglish: Ground state -13.6 eV hai. 12.5 eV wala electron atom ko -13.6 + 12.5 = -1.1 eV wale level tak le ja sakta hai. -13.6/n² se E3 = -1.51 eV aur E4 = -0.85 eV; -1.1 eV inke beech hai, isliye excitation n = 3 tak possible hai, n = 4 tak nahi. Iske baad n = 2 aur n = 3 se neeche transitions hote hain, isliye Lyman (n = 1 tak) aur Balmer (n = 2 tak) series emit hoti hain; n = 4 tak pahunch nahi sakta, isliye higher series ki lines nahi aayengi.

To check how much of this stuck, take the five-question quiz below. Each answer explains the reasoning, so read the explanation even when you get it right.

Quiz: the five question check for this chapter is set from the same topics. Attempt it after the questions above, then revisit any topic page where you lost a mark.