Structure of Atom Class 11 Chemistry NCERT Solutions

Step-by-step solution

1. 1(i) Mass of one electron = 9.10939 × 10⁻³¹ kg. The number of electrons weighing 1 g = 10⁻³ kg is (1 × 10⁻³)/(9.10939 × 10⁻³¹).
2. 2
   $$n=\frac{1\times 10^{-3}\text{kg}}{9.10939\times 10^{-31}\text{kg}}=1.098\times 10^{27}\text{electrons}$$
3. 3(ii) Mass of one mole of electrons = N_A × (mass of one electron) = (6.022 × 10²³)(9.10939 × 10⁻³¹ kg).
4. 4
   $$m=6.022\times 10^{23}\times 9.10939\times 10^{-31}=5.48\times 10^{-7}\text{kg}$$
5. 5Charge on one electron = 1.6022 × 10⁻¹⁹ C, so the charge on one mole of electrons = (1.6022 × 10⁻¹⁹)(6.022 × 10²³).
6. 6
   $$q=1.6022\times 10^{-19}\times 6.022\times 10^{23}=9.65\times 10^4\text{C}$$

Step-by-step solution

1. 1(i) One CH₄ molecule has 6 electrons of carbon and 4 electrons of hydrogen = 10 electrons. One mole = 6.023 × 10²³ molecules, so electrons = 10 × 6.023 × 10²³.
2. 2
   $$\text{electrons}=10\times 6.023\times 10^{23}=6.023\times 10^{24}$$
3. 3(ii) ¹⁴C has mass number 14 and atomic number 6, so neutrons per atom = 14 − 6 = 8. Atoms in 7 mg = (6.023 × 10²³ × 7 × 10⁻³)/14.
4. 4
   $$(a)\text{neutrons}=\frac{6.023\times 10^{23}\times 8}{14}\times 7\times 10^{-3}=2.4088\times 10^{21}$$
5. 5
   $$(b)\text{mass}=2.4088\times 10^{21}\times 1.675\times 10^{-27}=4.035\times 10^{-6}\text{kg}$$
6. 6(iii) One NH₃ molecule has 7 protons (nitrogen) + 3 protons (hydrogen) = 10 protons. Molecules in 34 mg of NH₃ (M = 17 g mol⁻¹) = (6.023 × 10²³ × 34 × 10⁻³)/17.
7. 7
   $$(a)\text{molecules}=\frac{6.023\times 10^{23}\times 34\times 10^{-3}}{17}=1.2044\times 10^{21}$$
8. 8
   $$\text{protons}=10\times 1.2044\times 10^{21}=1.2044\times 10^{22}$$
9. 9
   $$(b)\text{mass}=1.2044\times 10^{22}\times 1.6726\times 10^{-27}=2.0138\times 10^{-5}\text{kg}$$
10. 10The numbers and masses depend only on the mole–atom relationships, not on temperature or pressure — so the answers do NOT change with temperature and pressure.

Step-by-step solution

1. 1For a nucleus AZX: protons = Z (atomic number), neutrons = A − Z (mass number − atomic number).
2. 2₆¹³C: protons = 6, neutrons = 13 − 6 = 7.
3. 3₈¹⁶O: protons = 8, neutrons = 16 − 8 = 8.
4. 4₁₂²⁴Mg: protons = 12, neutrons = 24 − 12 = 12.
5. 5₂₆⁵⁶Fe: protons = 26, neutrons = 56 − 26 = 30.
6. 6₃₈⁸⁸Sr: protons = 38, neutrons = 88 − 38 = 50.

Step-by-step solution

1. 1The complete symbol records the atomic number as the subscript and the mass number as the superscript: AZX.
2. 2(i) Z = 17 is chlorine and A = 35, so the symbol is ₁₇³⁵Cl.
3. 3(ii) Z = 92 is uranium and A = 233, so the symbol is ₉₂²³³U.
4. 4(iii) Z = 4 is beryllium and A = 9, so the symbol is ₄⁹Be.

Step-by-step solution

1. 1Using c = νλ, the frequency is ν = c/λ, with c = 3 × 10⁸ m s⁻¹ and λ = 580 nm = 580 × 10⁻⁹ m.
2. 2
   $$\nu =\frac{3\times 10^8{\text{m s}}^{-1}}{580\times 10^{-9}\text{m}}=5.17\times 10^{14}{\text{s}}^{-1}$$
3. 3The wavenumber is the reciprocal of the wavelength: ν̄ = 1/λ.
4. 4
   $$\bar{\nu }=\frac{1}{580\times 10^{-9}\text{m}}=1.72\times 10^6{\text{m}}^{-1}$$

Step-by-step solution

1. 1(i) Energy of a photon: E = hν, with h = 6.626 × 10⁻³⁴ J s and ν = 3 × 10¹⁵ Hz.
2. 2
   $$E=(6.626\times 10^{-34}\text{J s})(3\times 10^{15}{\text{s}}^{-1})=1.988\times 10^{-18}\text{J}$$
3. 3(ii) Energy from wavelength: E = hc/λ, with λ = 0.50 Å = 0.50 × 10⁻¹⁰ m and c = 3 × 10⁸ m s⁻¹.
4. 4
   $$E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{0.50\times 10^{-10}}=3.976\times 10^{-15}\text{J}\approx 3.98\times 10^{-15}\text{J}$$

Step-by-step solution

1. 1Frequency is the reciprocal of the period: ν = 1/T.
2. 2
   $$\nu =\frac{1}{2.0\times 10^{-10}\text{s}}=5\times 10^9{\text{s}}^{-1}$$
3. 3Wavelength from c = νλ: λ = c/ν.
4. 4
   $$\lambda =\frac{3\times 10^8{\text{m s}}^{-1}}{5\times 10^9{\text{s}}^{-1}}=6.0\times 10^{-2}\text{m}$$
5. 5Wavenumber: ν̄ = 1/λ.
6. 6
   $$\bar{\nu }=\frac{1}{6.0\times 10^{-2}\text{m}}=16.66{\text{m}}^{-1}$$

Step-by-step solution

1. 1Energy of n photons of wavelength λ: E_n = n × (hc/λ), so n = E_n × λ/(hc).
2. 2Here E_n = 1 J, λ = 4000 pm = 4000 × 10⁻¹² m, c = 3 × 10⁸ m s⁻¹, h = 6.626 × 10⁻³⁴ J s.
3. 3
   $$n=\frac{1\times 4000\times 10^{-12}}{(6.626\times 10^{-34})(3\times 10^8)}=2.012\times 10^{16}$$

Step-by-step solution

1. 1(i) Photon energy E = hc/λ = (6.626 × 10⁻³⁴)(3 × 10⁸)/(4 × 10⁻⁷).
2. 2
   $$E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{4\times 10^{-7}}=4.9695\times 10^{-19}\text{J}$$
3. 3
   $$E=\frac{4.9695\times 10^{-19}}{1.6020\times 10^{-19}}=3.10\text{eV}$$
4. 4(ii) Einstein's photoelectric equation: KE = hν − W₀ = (3.10 − 2.13) eV.
5. 5
   $$KE=3.1020-2.13=0.97\text{eV}=1.554\times 10^{-19}\text{J}$$
6. 6(iii) From ½mv² = KE with m = 9.10939 × 10⁻³¹ kg:
7. 7
   $$v=\sqrt{\frac{2\times 0.9720\times 1.6020\times 10^{-19}}{9.10939\times 10^{-31}}}=\sqrt{0.3418\times 10^{12}}=5.84\times 10^5{\text{m s}}^{-1}$$

Step-by-step solution

1. 1Ionisation energy per mole: E = N_A hc/λ, with N_A = 6.023 × 10²³ mol⁻¹, h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, λ = 242 × 10⁻⁹ m.
2. 2
   $$E=\frac{(6.023\times 10^{23})(6.626\times 10^{-34})(3\times 10^8)}{242\times 10^{-9}}=4.947\times 10^5{\text{J mol}}^{-1}$$
3. 3
   $$E=494.7\times 10^3{\text{J mol}}^{-1}=494{\text{kJ mol}}^{-1}$$

Step-by-step solution

1. 1Power P = 25 W = 25 J s⁻¹. Energy of one photon: E = hc/λ with λ = 0.57 × 10⁻⁶ m.
2. 2
   $$E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{0.57\times 10^{-6}}=34.87\times 10^{-20}\text{J}$$
3. 3Number of quanta emitted per second = P/E.
4. 4
   $$\text{Rate}=\frac{25{\text{J s}}^{-1}}{34.87\times 10^{-20}\text{J}}=7.17\times 10^{19}{\text{s}}^{-1}$$

Step-by-step solution

1. 1Zero-velocity emission means the radiation just equals the threshold, λ₀ = 6800 Å = 6800 × 10⁻¹⁰ m = 6.8 × 10⁻⁷ m.
2. 2
   $${\nu }_0=\frac{c}{{\lambda }_0}=\frac{3\times 10^8{\text{m s}}^{-1}}{6.8\times 10^{-7}\text{m}}=4.41\times 10^{14}{\text{s}}^{-1}$$
3. 3Work function W₀ = hν₀.
4. 4
   $$W_0=(6.626\times 10^{-34})(4.41\times 10^{14})=2.92\times 10^{-19}\text{J}$$

Step-by-step solution

1. 1The n = 4 → n = 2 transition gives a spectral line of the Balmer series. The energy change is:
2. 2
   $$\Delta E=2.18\times 10^{-18}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)=2.18\times 10^{-18}\left(\frac{1}{4^2}-\frac{1}{2^2}\right)$$
3. 3
   $$\Delta E=2.18\times 10^{-18}\times \left(-\frac{3}{16}\right)=-4.0875\times 10^{-19}\text{J}$$
4. 4The negative sign shows the energy is emitted. The wavelength follows from E = hc/λ:
5. 5
   $$\lambda =\frac{hc}{E}=\frac{(6.626\times 10^{-34})(3\times 10^8)}{4.0875\times 10^{-19}}=4.863\times 10^{-7}\text{m}\approx 486\text{nm}$$

Step-by-step solution

1. 1Bohr energy: E_n = −(2.18 × 10⁻¹⁸)Z²/n² J, with Z = 1 for hydrogen. Ionisation from n₁ = 5 to n₂ = ∞:
2. 2
   $$\Delta E=E_{\infty }-E_5=\left(0\right)-\left(-\frac{2.18\times 10^{-18}}{25}\right)=\frac{2.18\times 10^{-18}}{25}=8.72\times 10^{-20}\text{J}$$
3. 3Ionisation from the ground state, n₁ = 1 to n₂ = ∞:
4. 4
   $$\Delta E^{\prime }=E_{\infty }-E_1=2.18\times 10^{-18}\text{J}$$
5. 58.72 × 10⁻²⁰ J is far smaller than 2.18 × 10⁻¹⁸ J — the electron in n = 5 is much easier to remove than one in the ground state.

Step-by-step solution

1. 1All allowed downward transitions 6→5, 6→4, 6→3, 6→2, 6→1, 5→4, …, 2→1 can occur.
2. 2Number of spectral lines when the electron in the n-th level drops to the ground state:
3. 3
   $$\text{Lines}=\frac{n(n-1)}{2}=\frac{6\times 5}{2}=15$$
4. 4That is 5 + 4 + 3 + 2 + 1 = 15 possible transitions.

Step-by-step solution

1. 1(i) E_n = E₁/n², so E₅ = −2.18 × 10⁻¹⁸/5².
2. 2
   $$E_5=\frac{-2.18\times 10^{-18}}{25}=-8.72\times 10^{-20}\text{J}$$
3. 3(ii) Bohr radius: r_n = (0.0529 nm) × n², with n = 5.
4. 4
   $$r_5=0.0529\times 25=1.3225\text{nm}$$

Step-by-step solution

1. 1For the Balmer series the lower level is n_i = 2. The wavenumber is ν̄ = R_H(1/2² − 1/n_f²) with R_H = 1.097 × 10⁷ m⁻¹.
2. 2Longest wavelength means smallest wavenumber, which arises from the smallest n_f allowed, n_f = 3:
3. 3
   $$\bar{\nu }=1.097\times 10^7\left(\frac{1}{4}-\frac{1}{9}\right)=1.097\times 10^7\times \frac{5}{36}$$
4. 4
   $$\bar{\nu }=1.5236\times 10^6{\text{m}}^{-1}$$

Step-by-step solution

1. 1Convert the ground-state energy to joules: 1 erg = 10⁻⁷ J, so E₁ = −2.18 × 10⁻¹¹ × 10⁻⁷ J = −2.18 × 10⁻¹⁸ J.
2. 2Energy required to go from n = 1 to n = 5: ΔE = E₅ − E₁.
3. 3
   $$\Delta E=-\frac{2.18\times 10^{-18}}{25}-(-2.18\times 10^{-18})=(2.18\times 10^{-18})\left(1-\frac{1}{25}\right)$$
4. 4
   $$\Delta E=2.18\times 10^{-18}\times \frac{24}{25}=2.0928\times 10^{-18}\text{J}$$
5. 5On returning to the ground state the same energy is emitted as light: λ = hc/ΔE.
6. 6
   $$\lambda =\frac{(6.626\times 10^{-34})(3\times 10^8)}{2.0928\times 10^{-18}}=9.498\times 10^{-8}\text{m}$$

Step-by-step solution

1. 1Removing the electron means ionising from n = 2 to n = ∞:
2. 2
   $$\Delta E=E_{\infty }-E_2=0-\left(-\frac{2.18\times 10^{-18}}{4}\right)=5.45\times 10^{-19}\text{J}$$
3. 3The longest wavelength that supplies this energy is λ = hc/ΔE.
4. 4
   $$\lambda =\frac{(6.626\times 10^{-34})(3\times 10^8)}{5.45\times 10^{-19}}=3.647\times 10^{-7}\text{m}$$
5. 5
   $$\lambda =3.647\times 10^{-7}\times 10^2\text{cm}=3.65\times 10^{-5}\text{cm}$$

Step-by-step solution

1. 1de Broglie's equation: λ = h/mv, with m = 9.10939 × 10⁻³¹ kg and v = 2.05 × 10⁷ m s⁻¹.
2. 2
   $$\lambda =\frac{6.626\times 10^{-34}\text{J s}}{(9.10939\times 10^{-31}\text{kg})(2.05\times 10^7{\text{m s}}^{-1})}$$
3. 3
   $$\lambda =3.548\times 10^{-11}\text{m}$$

Step-by-step solution

1. 1From KE = ½mv², the velocity is v = √(2 KE/m).
2. 2
   $$v=\sqrt{\frac{2\times 3.0\times 10^{-25}}{9.10939\times 10^{-31}}}=\sqrt{6.5866\times 10^5}=811.6{\text{m s}}^{-1}$$
3. 3Then λ = h/mv.
4. 4
   $$\lambda =\frac{6.626\times 10^{-34}}{(9.10939\times 10^{-31})(811.6)}=8.9625\times 10^{-7}\text{m}$$

Step-by-step solution

1. 1Isoelectronic species have the same electron count. Na (Z = 11) loses one electron to give Na⁺ with 10 electrons; Mg (Z = 12) loses two to give Mg²⁺ with 10.
2. 2K (Z = 19) loses one → K⁺ = 18 electrons; Ca (Z = 20) loses two → Ca²⁺ = 18; S (Z = 16) gains two → S²⁻ = 18; neutral Ar (Z = 18) has 18 electrons.
3. 3Group (i): Na⁺ and Mg²⁺ both have 10 electrons.
4. 4Group (ii): K⁺, Ca²⁺, S²⁻ and Ar all have 18 electrons.

Step-by-step solution

1. 1(i)(a) H atom is 1s¹; gaining one electron gives H⁻ = 1s².
2. 2(i)(b) Na atom is 1s²2s²2p⁶3s¹; losing one electron gives Na⁺ = 1s²2s²2p⁶.
3. 3(i)(c) O atom is 1s²2s²2p⁴; gaining two electrons gives O²⁻ = 1s²2s²2p⁶.
4. 4(i)(d) F atom is 1s²2s²2p⁵; gaining one electron gives F⁻ = 1s²2s²2p⁶.
5. 5(ii)(a) 3s¹: complete configuration 1s²2s²2p⁶3s¹ → 2 + 2 + 6 + 1 = 11 electrons, Z = 11.
6. 6(ii)(b) 2p³: complete configuration 1s²2s²2p³ → 2 + 2 + 3 = 7 electrons, Z = 7.
7. 7(ii)(c) 3p⁵: complete configuration 1s²2s²2p⁶3s²3p⁵ → 2 + 2 + 6 + 2 + 5 = 17 electrons, Z = 17.
8. 8(iii)(a) [He]2s¹ = 1s²2s¹ → Z = 3, lithium (Li).
9. 9(iii)(b) [Ne]3s²3p³ → Z = 15, phosphorus (P).
10. 10(iii)(c) [Ar]4s²3d¹ → Z = 21, scandium (Sc).

Step-by-step solution

1. 1The azimuthal quantum number takes values l = 0 to (n − 1); the g orbital has l = 4.
2. 2For l = 4 to be allowed, n − 1 ≥ 4, i.e. n ≥ 5.

Step-by-step solution

1. 1For a 3d orbital: the principal quantum number n = 3.
2. 2The d orbital has azimuthal quantum number l = 2.
3. 3The magnetic quantum number takes the (2l + 1) = 5 values m_l = −2, −1, 0, 1, 2.

Step-by-step solution

1. 1(i) In a neutral atom the number of protons equals the number of electrons, so the element has 29 protons (Z = 29, copper).
2. 2(ii) Filling in Aufbau order, 29 electrons give 1s²2s²2p⁶3s²3p⁶4s²3d⁹. But a filled 3d subshell is unusually stable, so one electron transfers from 4s to 3d:
3. 3Final configuration: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹ (the famous copper anomaly — a full 3d¹⁰ shell with a half-full 4s¹).

Step-by-step solution

1. 1H₂: each hydrogen contributes 1 electron, so H₂ has 1 + 1 = 2 electrons.
2. 2H₂⁺ loses one electron (positive charge = loss of an electron): electrons = 2 − 1 = 1.
3. 3O₂: each oxygen (Z = 8) contributes 8 electrons, so O₂ has 8 + 8 = 16 electrons.
4. 4O₂⁺ loses one electron: electrons = 16 − 1 = 15.

Step-by-step solution

1. 1(i) For n = 3, l takes values 0 to (n − 1) = 0, 1, 2.
2. 2For each l, m_l runs from −l to +l: l = 0 → m_l = 0; l = 1 → m_l = −1, 0, 1; l = 2 → m_l = −2, −1, 0, 1, 2.
3. 3(ii) For the 3d orbital l = 2 and there are (2l + 1) = 5 values: m_l = −2, −1, 0, 1, 2.
4. 4(iii) A p orbital needs l = 1, which requires n ≥ 2, so 1p is not possible but 2p and 2s are possible.
5. 5An f orbital needs l = 3, which requires n ≥ 4, so 3f is not possible.

Step-by-step solution

1. 1The orbital name is written n followed by the letter for l (0→s, 1→p, 2→d, 3→f).
2. 2(a) n = 1, l = 0 → 1s.
3. 3(b) n = 3, l = 1 → 3p.
4. 4(c) n = 4, l = 2 → 4d.
5. 5(d) n = 4, l = 3 → 4f.

Step-by-step solution

1. 1(a) Not possible: the principal quantum number n must be a positive integer, so n = 0 is not allowed.
2. 2(b) Possible: n = 1, l = 0, m_l = 0 and m_s = ±1/2 is the valid description of a 1s electron.
3. 3(c) Not possible: for n = 1, l can only be 0 (values 0 to n − 1), so l = 1 is not allowed.
4. 4(d) Possible: n = 2, l = 1, m_l = 0, m_s = −1/2 correctly describes a 2p electron.
5. 5(e) Not possible: for n = 3, l can be only 0, 1, 2 — l = 3 exceeds n − 1.
6. 6(f) Possible: n = 3, l = 1, m_l = 0, m_s = +1/2 correctly describes a 3p electron.

Step-by-step solution

1. 1(a) A shell with quantum number n can hold at most 2n² electrons; for n = 4 that is 2 × 16 = 32 electrons.
2. 2Exactly half the electrons of a filled shell have m_s = −1/2, so the number is 32/2 = 16.
3. 3(b) n = 3, l = 0 is the 3s orbital, which holds a maximum of 2 electrons.

Step-by-step solution

1. 1Bohr's quantisation of angular momentum: mvr = n(h/2π), with n = 1, 2, 3, ….
2. 2de Broglie's equation: λ = h/mv, so that mv = h/λ.
3. 3Substitute mv = h/λ into the Bohr condition:
4. 4
   $$\frac{h}{\lambda }r=n\frac{h}{2\pi }\Rightarrow 2\pi r=n\lambda$$
5. 5Since 2πr is the circumference of the orbit, the circumference is exactly n times the de Broglie wavelength — an integral multiple — as required for a standing matter wave.

Step-by-step solution

1. 1For a hydrogen-like species the wavenumber is ν̄ = (1/λ) = R Z²(1/n₁² − 1/n₂²), with Z = 2 for helium.
2. 2For He⁺ with n₁ = 2 and n₂ = 4:
3. 3
   $$\bar{\nu }=R\times 2^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=4R\times \frac{3}{16}=\frac{3R}{4}$$
4. 4The hydrogen transition (Z = 1) must give the same wavenumber: (1/n₁² − 1/n₂²) = 3/4.
5. 5By inspection, n₁ = 1 and n₂ = 2 give 1 − 1/4 = 3/4, the only matching pair.

Step-by-step solution

1. 1Energy levels of a hydrogen-like species: E_n = −2.18 × 10⁻¹⁸ Z²/n² J.
2. 2For hydrogen (Z = 1), ionisation energy = 0 − E₁ = 2.18 × 10⁻¹⁸ J, confirming the given value.
3. 3For the process He⁺ → He²⁺ + e⁻, Z = 2 and the electron is removed from n = 1:
4. 4
   $$\Delta E=E_{\infty }-E_1=2.18\times 10^{-18}\times \frac{Z^2}{n^2}=2.18\times 10^{-18}\times 4=8.72\times 10^{-18}\text{J}$$

Step-by-step solution

1. 1Convert the length of the scale: 20 cm = 20 × 10⁻² m = 0.2 m.
2. 2Diameter of one carbon atom = 0.15 nm = 0.15 × 10⁻⁹ m.
3. 3Number of atoms = length/diameter:
4. 4
   $$N=\frac{20\times 10^{-2}\text{m}}{0.15\times 10^{-9}\text{m}}=133.33\times 10^7=1.33\times 10^9$$

Step-by-step solution

1. 1Length of the arrangement = 2.4 cm = 2.4 × 10⁻² m.
2. 2Diameter of one carbon atom = length/number of atoms = (2.4 × 10⁻²)/(2 × 10⁸).
3. 3
   $$\text{Diameter}=1.2\times 10^{-10}\text{m}$$
4. 4Radius = diameter/2:
5. 5
   $$r=\frac{1.2\times 10^{-10}}{2}=6.0\times 10^{-11}\text{m}$$

Step-by-step solution

1. 1(a) Radius = diameter/2 = 2.6/2 = 1.3 Å = 1.3 × 10⁻¹⁰ m.
2. 2
   $$r=1.3\times 10^{-10}\text{m}=130\times 10^{-12}\text{m}=130\text{pm}$$
3. 3(b) Length = 1.6 cm = 1.6 × 10⁻² m; diameter = 2.6 × 10⁻¹⁰ m. Number of atoms = length/diameter.
4. 4
   $$N=\frac{1.6\times 10^{-2}}{2.6\times 10^{-10}}=6.15\times 10^7$$

Step-by-step solution

1. 1Charge on one electron = 1.6022 × 10⁻¹⁹ C.
2. 2Number of electrons = total charge/charge on one electron.
3. 3
   $$n=\frac{2.5\times 10^{-16}\text{C}}{1.6022\times 10^{-19}\text{C}}=1.560\times 10^3\approx 1560$$

Step-by-step solution

1. 1Charge on the oil drop = 1.282 × 10⁻¹⁸ C; charge on one electron = 1.6022 × 10⁻¹⁹ C.
2. 2
   $$n=\frac{1.282\times 10^{-18}\text{C}}{1.6022\times 10^{-19}\text{C}}=0.8001\times 10^1=8.0$$

Step-by-step solution

1. 1The large-angle scattering of α-particles comes from the strong positive charge concentrated in a heavy nucleus.
2. 2A light atom carries very little positive charge in its nucleus, so the repulsive force on the positively charged α-particle is weak.
3. 3Result: far fewer particles are deflected, and the big-angle deflections almost disappear — the scattering pattern is much closer to the straight-through path.

Step-by-step solution

1. 1The convention is to write Z as the subscript and A as the superscript: AZX.
2. 2So ₃₅⁷⁹Br is the full symbol, and ⁷⁹Br (mass number shown, atomic number elided) is also acceptable because the element symbol already fixes Z.
3. 3₇₉³⁵Br would put the atomic number where the mass number belongs, which is incorrect.
4. 4³⁵Br is not acceptable because the atomic number of an element is fixed while the atomic mass varies with the isotope — citing only a number below the symbol implies Z = 35, which is bromine; but a superscript 35 would be a misleading mass.

Step-by-step solution

1. 1Let the number of protons be x. Then the number of neutrons = x + 0.317x = 1.317x.
2. 2Mass number = protons + neutrons = x + 1.317x = 2.317x = 81.
3. 3
   $$x=\frac{81}{2.317}=34.95\approx 35$$
4. 4Z = 35 (bromine) and A = 81, so the symbol is ₃₅⁸¹Br.

Step-by-step solution

1. 1Let x be the number of electrons in the ion. Since it is a 1− anion, it has one electron more than protons, so protons = x − 1.
2. 2Neutrons = x + 0.111x = 1.111x.
3. 3Mass number = protons + neutrons = (x − 1) + 1.111x = 2.111x − 1 = 37.
4. 4
   $$2.111x=38\Rightarrow x=18\text{electrons}$$
5. 5Protons = x − 1 = 17, which is chlorine (Z = 17), with mass number 37.

Step-by-step solution

1. 1Let x be the number of electrons in the ion A³⁺. The ion has lost 3 electrons, so the neutral atom has x + 3 electrons = x + 3 protons.
2. 2Neutrons = x + 0.304x = 1.304x.
3. 3Mass number = protons + neutrons = (x + 3) + 1.304x = 2.304x + 3 = 56.
4. 4
   $$2.304x=53\Rightarrow x=23\text{electrons}$$
5. 5Protons = x + 3 = 26, which is iron (Z = 26), with mass number 56.

Step-by-step solution

1. 1Recall the approximate frequency ranges: FM radio ~10⁸ Hz, microwave oven ~10¹⁰ Hz, visible (amber) light ~5 × 10¹⁴ Hz, X-rays ~10¹⁸ Hz, cosmic rays ~10²⁰ Hz and above.
2. 2Increasing frequency: FM radio < microwave oven < amber light < X-rays < cosmic rays.
3. 3Wavelength is the reverse of this order: cosmic rays < X-rays < microwave oven < amber light < FM radio.

Step-by-step solution

1. 1Total energy emitted = N × (energy per photon) = N hc/λ.
2. 2With N = 5.6 × 10²⁴, h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, λ = 337.1 × 10⁻⁹ m:
3. 3
   $$E=\frac{(5.6\times 10^{24})(6.626\times 10^{-34})(3\times 10^8)}{337.1\times 10^{-9}}=0.3302\times 10^7\text{J}$$
4. 4
   $$E=3.33\times 10^6\text{J}$$

Step-by-step solution

1. 1(a) ν = c/λ with λ = 616 × 10⁻⁹ m.
2. 2
   $$\nu =\frac{3.0\times 10^8{\text{m s}}^{-1}}{616\times 10^{-9}\text{m}}=4.87\times 10^{14}{\text{s}}^{-1}$$
3. 3(b) Distance = speed × time = (3.0 × 10⁸ m s⁻¹)(30 s).
4. 4
   $$d=9.0\times 10^9\text{m}$$
5. 5(c) Energy of one quantum E = hν = (6.626 × 10⁻³⁴)(4.87 × 10¹⁴).
6. 6
   $$E=32.27\times 10^{-20}\text{J}=3.227\times 10^{-19}\text{J}$$
7. 7(d) Number of quanta in 2 J of energy = 2/E.
8. 8
   $$n=\frac{2\text{J}}{3.227\times 10^{-19}\text{J}}=6.2\times 10^{18}$$

Step-by-step solution

1. 1Energy of one photon: E = hc/λ with λ = 600 × 10⁻⁹ m.
2. 2
   $$E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{600\times 10^{-9}}=3.313\times 10^{-19}\text{J}$$
3. 3Number of photons = total energy/energy per photon.
4. 4
   $$n=\frac{3.15\times 10^{-18}}{3.313\times 10^{-19}}=9.5\approx 10$$

Step-by-step solution

1. 1Frequency of the radiation: ν = 1/T = 1/(2 × 10⁻⁹ s).
2. 2
   $$\nu =5\times 10^8{\text{s}}^{-1}$$
3. 3Energy of the source: E = N h ν with N = 2.5 × 10¹⁵.
4. 4
   $$E=(2.5\times 10^{15})(6.626\times 10^{-34})(5\times 10^8)=8.28\times 10^{-10}\text{J}$$

Step-by-step solution

1. 1For λ₁ = 589 nm: ν₁ = c/λ₁.
2. 2
   $${\nu }_1=\frac{3.0\times 10^8}{589\times 10^{-9}}=5.093\times 10^{14}{\text{s}}^{-1}$$
3. 3For λ₂ = 589.6 nm: ν₂ = c/λ₂.
4. 4
   $${\nu }_2=\frac{3.0\times 10^8}{589.6\times 10^{-9}}=5.088\times 10^{14}{\text{s}}^{-1}$$
5. 5Energy difference between the two excited states: ΔE = h(ν₁ − ν₂).
6. 6
   $$\Delta E=(6.626\times 10^{-34})(5.093-5.088)\times 10^{14}=3.31\times 10^{-22}\text{J}$$

Step-by-step solution

1. 1(a) From W₀ = hc/λ₀, the threshold wavelength is λ₀ = hc/W₀, with W₀ = 1.9 × 1.602 × 10⁻¹⁹ J.
2. 2
   $${\lambda }_0=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{1.9\times 1.602\times 10^{-19}}=6.53\times 10^{-7}\text{m}=653\text{nm}$$
3. 3(b) Threshold frequency ν₀ = W₀/h.
4. 4
   $${\nu }_0=\frac{1.9\times 1.602\times 10^{-19}}{6.626\times 10^{-34}}=4.593\times 10^{14}{\text{s}}^{-1}$$
5. 5(c) Kinetic energy at λ = 500 nm: KE = hc(1/λ − 1/λ₀).
6. 6
   $$KE=(1.9878\times 10^{-25}\text{J m})\left(\frac{1}{500\times 10^{-9}}-\frac{1}{653\times 10^{-9}}\right)=9.31\times 10^{-20}\text{J}$$
7. 7Velocity from KE = ½mv² with m = 9.10939 × 10⁻³¹ kg:
8. 8
   $$v=\sqrt{\frac{2\times 9.31\times 10^{-20}}{9.10939\times 10^{-31}}}=4.52\times 10^5{\text{m s}}^{-1}$$

Step-by-step solution

1. 1Kinetic energy of the ejected electron: h(ν − ν₀) = ½mv². In terms of wavelengths, hc(1/λ − 1/λ₀) = ½mv².
2. 2Convert velocities: 2.55 × 10³ m s⁻¹, 4.35 × 10³ m s⁻¹, 5.35 × 10³ m s⁻¹. Write the relations for λ = 400 nm and λ = 500 nm and divide:
3. 3
   $$\frac{1/400-1/{\lambda }_0}{1/500-1/{\lambda }_0}={\left(\frac{5.35}{2.55}\right)}^2=4.40177$$
4. 4Simplify: 5(λ₀ − 400) = 4.40177 × 4(λ₀ − 500).
5. 5
   $$17.607{\lambda }_0-5{\lambda }_0=8803.5-2000\Rightarrow {\lambda }_0=\frac{6803.5}{12.607}=539.8\text{nm}$$
6. 6(a) Threshold wavelength λ₀ ≈ 540 nm.
7. 7(b) The tabulated velocities are not consistent — substituting them into hc(1/λ − 1/λ₀) = ½mv² does not reproduce Planck's constant (the data appear to be mis-scaled). For this reason Planck's constant cannot be reliably extracted from the stated figures.

Step-by-step solution

1. 1Conservation of energy: E = W₀ + KE, so W₀ = E − KE. The stopping voltage gives KE = e × 0.35 V = 0.35 eV.
2. 2Energy of the incident photon: E = hc/λ with λ = 256.7 × 10⁻⁹ m.
3. 3
   $$E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{256.7\times 10^{-9}}=7.744\times 10^{-19}\text{J}=4.83\text{eV}$$
4. 4Work function: W₀ = 4.83 − 0.35.
5. 5
   $$W_0=4.48\text{eV}$$

Step-by-step solution

1. 1Energy of the incident photon: E = hc/λ with λ = 150 × 10⁻¹² m.
2. 2
   $$E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{150\times 10^{-12}}=1.3252\times 10^{-15}\text{J}=13.252\times 10^{-16}\text{J}$$
3. 3Kinetic energy of the ejected electron: KE = ½m_e v².
4. 4
   $$KE=\frac{1}{2}(9.10939\times 10^{-31})(1.5\times 10^7{)}^2=1.025\times 10^{-16}\text{J}$$
5. 5Binding energy = E − KE.
6. 6
   $$E_b=(13.252-1.025)\times 10^{-16}=12.227\times 10^{-16}\text{J}=\frac{12.227\times 10^{-16}}{1.602\times 10^{-19}}\text{eV}$$
7. 7
   $$E_b=7.63\times 10^3\text{eV}$$

Step-by-step solution

1. 1The observed frequency is ν = c/λ = (3 × 10⁸)/(1285 × 10⁻⁹).
2. 2
   $$\nu =\frac{3\times 10^8}{1285\times 10^{-9}}=2.33\times 10^{14}{\text{s}}^{-1}$$
3. 3Set the given formula equal to this frequency:
4. 4
   $$3.29\times 10^{15}\left(\frac{1}{9}-\frac{1}{n^2}\right)=2.33\times 10^{14}$$
5. 5
   $$\frac{1}{9}-\frac{1}{n^2}=\frac{2.33\times 10^{14}}{3.29\times 10^{15}}=0.0708$$
6. 6
   $$\frac{1}{n^2}=0.1111-0.0708=0.0403\Rightarrow n=\sqrt{\frac{1}{0.0403}}=4.98\approx 5$$
7. 7n = 5 (Paschen series), and a wavelength of 1285 nm places the transition in the infrared region of the spectrum.

Step-by-step solution

1. 1Bohr radius for hydrogen: r = (52.9 n²/Z) pm with Z = 1, so n² = r/52.9 pm.
2. 2For r₁ = 1.3225 nm = 1322.5 pm: n₁² = 1322.5/52.9 = 25, so n₁ = 5.
3. 3For r₂ = 211.6 pm: n₂² = 211.6/52.9 = 4, so n₂ = 2.
4. 4The transition is 5 → 2, which belongs to the Balmer series. Its wavenumber:
5. 5
   $$\bar{\nu }=1.097\times 10^7\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=1.097\times 10^7\times \frac{21}{100}=2.303\times 10^6{\text{m}}^{-1}$$
6. 6Wavelength λ = 1/ν̄.
7. 7
   $$\lambda =\frac{1}{2.303\times 10^6}=4.34\times 10^{-7}\text{m}=434\text{nm}$$

Step-by-step solution

1. 1de Broglie's equation: λ = h/mv, with m = 9.10939 × 10⁻³¹ kg and v = 1.6 × 10⁶ m s⁻¹.
2. 2
   $$\lambda =\frac{6.626\times 10^{-34}}{(9.10939\times 10^{-31})(1.6\times 10^6)}=4.55\times 10^{-10}\text{m}$$
3. 3
   $$\lambda =4.55\times 10^{-10}\text{m}=455\text{pm}$$