Chapter 2: Structure of Atom (Structure of Atom)

(i) $n=\frac{1\times 10^{-3}\text{kg}}{9.109\times 10^{-31}\text{kg}}=1.098\times 10^{27}$ electrons.

(ii) Mass $=9.109\times 10^{-31}\times 6.022\times 10^{23}=5.48\times 10^{-7}$ kg; charge $=1.602\times 10^{-19}\times 6.022\times 10^{23}=9.65\times 10^4$ C.

(i) 1.098 × 10²⁷ (ii) 5.48 × 10⁻⁷ kg; 9.65 × 10⁴ C

(i) CH₄ has 6 + 4 = 10 electrons: $10\times 6.022\times 10^{23}=6.022\times 10^{24}$.

(ii) 7 mg ¹⁴C $=\frac{7\times 10^{-3}}{14}=5\times 10^{-4}$ mol; each atom has 14 − 6 = 8 neutrons. (a) $5\times 10^{-4}\times 8\times 6.022\times 10^{23}=2.409\times 10^{21}$ (b) $2.409\times 10^{21}\times 1.675\times 10^{-27}=4.035\times 10^{-6}$ kg.

(iii) 34 mg NH₃ $=\frac{34\times 10^{-3}}{17}=2\times 10^{-3}$ mol; each molecule has 7 + 3 = 10 protons. (a) $2\times 10^{-3}\times 10\times 6.022\times 10^{23}=1.204\times 10^{22}$ (b) $\times 1.6726\times 10^{-27}=2.014\times 10^{-5}$ kg.

No: the numbers depend only on the amount (mass) taken, not on temperature and pressure.

(i) 6.022 × 10²⁴ (ii) 2.409 × 10²¹; 4.035 × 10⁻⁶ kg (iii) 1.204 × 10²²; 2.014 × 10⁻⁵ kg; no change

Protons = Z (lower number), neutrons = A − Z.

C: 6 p, 7 n; O: 8 p, 8 n; Mg: 12 p, 12 n; Fe: 26 p, 30 n; Sr: 38 p, 50 n

(i) ${}_{17}^{35}$Cl (ii) ${}_{92}^{233}$U (iii) ${}_4^9$Be

$\nu =\frac{c}{\lambda }=\frac{3.0\times 10^8}{580\times 10^{-9}}=5.17\times 10^{14}{\text{s}}^{-1}$; $\bar{\nu }=\frac{1}{\lambda }=\frac{1}{580\times 10^{-9}}=1.72\times 10^6{\text{m}}^{-1}$.

ν = 5.17 × 10¹⁴ s⁻¹; ν̄ = 1.72 × 10⁶ m⁻¹

(i) $E=h\nu =6.626\times 10^{-34}\times 3\times 10^{15}=1.988\times 10^{-18}$ J.

(ii) $E=\frac{hc}{\lambda }=\frac{6.626\times 10^{-34}\times 3\times 10^8}{0.50\times 10^{-10}}=3.98\times 10^{-15}$ J.

(i) 1.988 × 10⁻¹⁸ J (ii) 3.98 × 10⁻¹⁵ J

$\nu =\frac{1}{T}=5.0\times 10^9{\text{s}}^{-1}$; $\lambda =\frac{c}{\nu }=\frac{3\times 10^8}{5\times 10^9}=6.0\times 10^{-2}$ m; $\bar{\nu }=\frac{1}{\lambda }=16.66{\text{m}}^{-1}$.

λ = 6.0 × 10⁻² m; ν = 5.0 × 10⁹ s⁻¹; ν̄ = 16.66 m⁻¹

$E=\frac{hc}{\lambda }=\frac{6.626\times 10^{-34}\times 3\times 10^8}{4000\times 10^{-12}}=4.97\times 10^{-17}$ J per photon; $n=\frac{1}{4.97\times 10^{-17}}$.

2.012 × 10¹⁶ photons

(i) $E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{4\times 10^{-7}}=4.97\times 10^{-19}$ J $=\frac{4.97\times 10^{-19}}{1.602\times 10^{-19}}=3.10$ eV.

(ii) K.E. $=3.10-2.13=0.97$ eV $=1.55\times 10^{-19}$ J.

(iii) $v=\sqrt{\frac{2\times 1.55\times 10^{-19}}{9.109\times 10^{-31}}}=5.84\times 10^5{\text{m s}}^{-1}$.

(i) 4.97 × 10⁻¹⁹ J = 3.10 eV (ii) 0.97 eV (iii) 5.84 × 10⁵ m s⁻¹

Per atom: $E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{242\times 10^{-9}}=8.21\times 10^{-19}$ J. Per mole: $\times 6.022\times 10^{23}=4.94\times 10^5$ J.

494 kJ mol⁻¹

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{0.57\times 10^{-6}}=3.49\times 10^{-19}$ J; quanta per second $=\frac{25}{3.49\times 10^{-19}}$.

7.18 × 10¹⁹ s⁻¹

Zero velocity means the photon energy just equals the work function.

${\nu }_0=\frac{3\times 10^8}{6800\times 10^{-10}}=4.41\times 10^{14}{\text{s}}^{-1}$; $W_0=h{\nu }_0=6.626\times 10^{-34}\times 4.41\times 10^{14}=2.92\times 10^{-19}$ J.

ν₀ = 4.41 × 10¹⁴ s⁻¹; W₀ = 2.92 × 10⁻¹⁹ J (≈ 1.82 eV)

$\Delta E=2.18\times 10^{-18}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=2.18\times 10^{-18}\times \frac{3}{16}=4.09\times 10^{-19}$ J.

$\lambda =\frac{hc}{\Delta E}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.09\times 10^{-19}}=486\times 10^{-9}$ m.

486 nm (Balmer series, blue-green)

$E_5=-\frac{2.18\times 10^{-18}}{25}=-8.72\times 10^{-20}$ J, so $8.72\times 10^{-20}$ J is needed. From n = 1: $2.18\times 10^{-18}$ J, which is $5^2=25$ times larger.

8.72 × 10⁻²⁰ J; ionisation from n = 1 needs 25 times more

Every pair of levels from 6 down to 1 gives a line: $\frac{n(n-1)}{2}=\frac{6\times 5}{2}$ (= 5 + 4 + 3 + 2 + 1).

15 lines

(i) $E_5=\frac{E_1}{5^2}=\frac{-2.18\times 10^{-18}}{25}=-8.72\times 10^{-20}$ J. (ii) $r_5=0.0529\times 5^2$ nm $=1.3225$ nm.

(i) −8.72 × 10⁻²⁰ J (ii) 1.3225 nm

Longest λ = smallest energy gap: $n_2=3\to n_1=2$.

$\bar{\nu }=1.097\times 10^7\left(\frac{1}{4}-\frac{1}{9}\right)=1.097\times 10^7\times \frac{5}{36}$ m⁻¹.

1.523 × 10⁶ m⁻¹ (λ = 656 nm, red)

$\Delta E=2.18\times 10^{-11}\left(1-\frac{1}{25}\right)=2.09\times 10^{-11}$ erg $=2.09\times 10^{-18}$ J.

$\lambda =\frac{hc}{\Delta E}=\frac{6.626\times 10^{-27}\text{erg s}\times 3\times 10^{10}{\text{cm s}}^{-1}}{2.09\times 10^{-11}\text{erg}}=9.50\times 10^{-6}$ cm.

2.09 × 10⁻¹⁸ J (NCERT key: 2.08 × 10⁻¹¹ erg); λ = 950 Å

$E=\frac{2.18\times 10^{-18}}{4}=5.45\times 10^{-19}$ J. Longest λ $=\frac{hc}{E}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{5.45\times 10^{-19}}=3.647\times 10^{-7}$ m.

5.45 × 10⁻¹⁹ J; 3.647 × 10⁻⁵ cm (3647 Å)

$\lambda =\frac{h}{mv}=\frac{6.626\times 10^{-34}}{9.109\times 10^{-31}\times 2.05\times 10^7}$

3.55 × 10⁻¹¹ m

$v=\sqrt{\frac{2\text{K.E.}}{m}}=\sqrt{\frac{2\times 3.0\times 10^{-25}}{9.1\times 10^{-31}}}=812{\text{m s}}^{-1}$; $\lambda =\frac{6.626\times 10^{-34}}{9.1\times 10^{-31}\times 812}=8.97\times 10^{-7}$ m.

8967 Å

Electrons: Na⁺ 10, K⁺ 18, Mg²⁺ 10, Ca²⁺ 18, S²⁻ 18, Ar 18.

Na⁺ and Mg²⁺ (10 e⁻); K⁺, Ca²⁺, S²⁻ and Ar (18 e⁻)

(i) (a) H⁻ (2 e⁻): 1s². (b), (c), (d) Na⁺, O²⁻, F⁻ (10 e⁻ each): 1s² 2s² 2p⁶.

(ii) (a) 1s² 2s² 2p⁶ 3s¹: Z = 11 (Na) (b) 1s² 2s² 2p³: Z = 7 (N) (c) [Ne]3s² 3p⁵: Z = 17 (Cl).

(iii) (a) Z = 3: Li (b) Z = 15: P (c) Z = 21: Sc.

(i) 1s²; 1s² 2s² 2p⁶ (Na⁺, O²⁻, F⁻) (ii) 11, 7, 17 (iii) Li, P, Sc

g orbitals have l = 4, and l ≤ n − 1, so n ≥ 5.

n = 5

n = 3, l = 2, mₗ = −2, −1, 0, +1 or +2 (any one)

(i) A neutral atom has protons = electrons = 29 (copper, A = 64). (ii) Cu: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (a completely filled 3d is extra stable).

(i) 29 (ii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

H₂⁺: 2 − 1 = 1; H₂: 2; O₂⁺: 16 − 1 = 15.

1, 2 and 15

(i) l = 0 (mₗ = 0); l = 1 (mₗ = −1, 0, +1); l = 2 (mₗ = −2, −1, 0, +1, +2).

(ii) l = 2; mₗ = −2, −1, 0, +1, +2.

(iii) 2s and 2p are possible. 1p needs l = 1 with n = 1 (impossible, l ≤ 0); 3f needs l = 3 with n = 3 (impossible, l ≤ 2).

(i) l = 0, 1, 2 with the mₗ above (ii) l = 2, mₗ = −2 … +2 (iii) 2s and 2p

l = 0, 1, 2, 3 correspond to s, p, d, f.

(a) 1s (b) 3p (c) 4d (d) 4f

(a) n cannot be 0. (c) For n = 1, l can only be 0. (e) For n = 3, l can be at most 2. (b), (d), (f) obey all rules.

(a), (c) and (e) are not possible.

(a) n = 4 has n² = 16 orbitals; each can hold one electron with mₛ = −½: 16. (b) n = 3, l = 0 is the 3s orbital: 2 electrons.

(a) 16 (b) 2

Bohr: $mvr=\frac{nh}{2\pi }$, so $2\pi r=\frac{nh}{mv}$. de Broglie: $\lambda =\frac{h}{mv}$.

Hence $2\pi r=n\lambda$: a whole number of waves fits round the orbit (a standing wave).

2πr = nλ. Shown.

$\bar{\nu }\propto Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$. For He⁺ (Z = 2): $4\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3}{4}$.

For H (Z = 1) we need $\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{3}{4}=1-\frac{1}{4}$: $n_1=1$, $n_2=2$.

n = 2 → n = 1 (first Lyman line)

For hydrogen-like species $E\propto Z^2$: $E=2.18\times 10^{-18}\times 2^2$.

8.72 × 10⁻¹⁸ J atom⁻¹

$n=\frac{0.20\text{m}}{0.15\times 10^{-9}\text{m}}$

1.33 × 10⁹ atoms

Diameter $=\frac{2.4\times 10^{-2}}{2\times 10^8}=1.2\times 10^{-10}$ m; radius $=0.6\times 10^{-10}$ m.

0.06 nm

(a) $r=1.3$ Å $=130$ pm. (b) $n=\frac{1.6\times 10^{-2}}{2.6\times 10^{-10}}=6.15\times 10^7$.

(a) 1.3 × 10² pm (b) 6.15 × 10⁷ atoms

$n=\frac{2.5\times 10^{-16}}{1.602\times 10^{-19}}$

1560 electrons

$n=\frac{1.282\times 10^{-18}}{1.602\times 10^{-19}}$

8 electrons

A light atom has a much smaller nuclear charge and mass. The repulsion on α-particles is weaker, so more of them pass undeflected, fewer are deflected, and those deflected turn through smaller angles. Hardly any bounce back (a light nucleus may itself recoil).

More α-particles pass straight through; fewer and smaller deflections.

The symbol Br itself fixes Z = 35, so ⁷⁹Br (mass number only) is enough, and ${}_{35}^{79}$Br is complete. In ${}_{79}^{35}$Br the lower number 79 contradicts Z = 35 and A < Z is impossible; ³⁵Br would mean A = Z = 35 (no neutrons), which no bromine isotope has. A is written above, Z below, and A > Z.

Z is fixed by the symbol; A goes on top, Z at the bottom, and A > Z.

Let protons = x; neutrons = 1.317x. $x+1.317x=81\Rightarrow x=34.96\approx 35$ (bromine); neutrons = 46.

${}_{35}^{81}$Br

Let electrons = x; protons = x − 1; neutrons = 1.111x. $(x-1)+1.111x=37\Rightarrow x=18$. Protons = 17 (Cl), neutrons = 20.

${}_{17}^{37}$Cl⁻

Let electrons = x; protons = x + 3; neutrons = 1.304x. $(x+3)+1.304x=56\Rightarrow x=23$. Protons = 26 (Fe), neutrons = 30.

${}_{26}^{56}$Fe³⁺

Frequency increases as wavelength decreases.

FM radio < microwave < amber light < X-rays < cosmic rays, i.e. (c) < (a) < (b) < (e) < (d)

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{337.1\times 10^{-9}}=5.90\times 10^{-19}$ J per photon. Total $=5.6\times 10^{24}\times 5.90\times 10^{-19}=3.3\times 10^6$ J (taking these photons as emitted per second, the power is $3.3\times 10^6$ W).

3.3 × 10⁶ J (per second)

(a) $\nu =\frac{3\times 10^8}{616\times 10^{-9}}=4.87\times 10^{14}{\text{s}}^{-1}$ (b) $d=ct=3\times 10^8\times 30=9.0\times 10^9$ m (c) $E=h\nu =6.626\times 10^{-34}\times 4.87\times 10^{14}=3.23\times 10^{-19}$ J (d) $n=\frac{2}{3.23\times 10^{-19}}=6.2\times 10^{18}$

(a) 4.87 × 10¹⁴ s⁻¹ (b) 9.0 × 10⁹ m (c) 3.23 × 10⁻¹⁹ J (d) 6.2 × 10¹⁸ quanta

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{600\times 10^{-9}}=3.313\times 10^{-19}$ J per photon; $n=\frac{3.15\times 10^{-18}}{3.313\times 10^{-19}}=9.5$, i.e. about 10 (photons come whole).

About 10 photons

Frequency $\nu =\frac{1}{2\times 10^{-9}}=5.0\times 10^8{\text{s}}^{-1}$. Energy $=Nh\nu =2.5\times 10^{15}\times 6.626\times 10^{-34}\times 5.0\times 10^8$.

8.28 × 10⁻¹⁰ J

${\nu }_1=\frac{3\times 10^8}{589\times 10^{-9}}=5.093\times 10^{14}{\text{s}}^{-1}$; ${\nu }_2=\frac{3\times 10^8}{589.6\times 10^{-9}}=5.088\times 10^{14}{\text{s}}^{-1}$.

$\Delta E=h({\nu }_1-{\nu }_2)=6.626\times 10^{-34}\times 5.2\times 10^{11}=3.45\times 10^{-22}$ J.

5.093 × 10¹⁴ and 5.088 × 10¹⁴ s⁻¹; ΔE ≈ 3.45 × 10⁻²² J

$W_0=1.9\times 1.602\times 10^{-19}=3.04\times 10^{-19}$ J.

(a) ${\lambda }_0=\frac{hc}{W_0}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{3.04\times 10^{-19}}=6.52\times 10^{-7}$ m $=652$ nm. (b) ${\nu }_0=\frac{W_0}{h}=4.59\times 10^{14}{\text{s}}^{-1}$.

At 500 nm: $\frac{hc}{\lambda }=3.976\times 10^{-19}$ J; K.E. $=3.976\times 10^{-19}-3.04\times 10^{-19}=9.3\times 10^{-20}$ J; $v=\sqrt{\frac{2\times 9.3\times 10^{-20}}{9.109\times 10^{-31}}}=4.52\times 10^5{\text{m s}}^{-1}$.

(a) 652 nm (b) 4.59 × 10¹⁴ s⁻¹; K.E. 9.29 × 10⁻²⁰ J; v = 4.52 × 10⁵ m s⁻¹

Taking the speeds as 2.55, 4.35, 5.35 × 10⁵ m s⁻¹ (as cm s⁻¹ the values would be impossibly small):

(a) $\frac{1}{2}m{v}^2=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right)$, so $v^2\propto \frac{1}{\lambda }-\frac{1}{{\lambda }_0}$. Using the first two readings: $\frac{2.55^2}{4.35^2}=0.3436=\frac{1/500-1/{\lambda }_0}{1/450-1/{\lambda }_0}\Rightarrow {\lambda }_0=530.9$ nm.

(b) $\frac{1}{2}m{v}^2=h\nu -W_0$, so h is the slope of K.E. against ν. K.E. at 500 nm $=\frac{1}{2}(9.109\times 10^{-31})(2.55\times 10^5{)}^2=2.96\times 10^{-20}$ J; at 400 nm $=1.304\times 10^{-19}$ J. Frequencies: $6.0\times 10^{14}$ and $7.5\times 10^{14}$ s⁻¹.

$h=\frac{1.304\times 10^{-19}-2.96\times 10^{-20}}{7.5\times 10^{14}-6.0\times 10^{14}}=6.7\times 10^{-34}$ J s, close to the accepted $6.626\times 10^{-34}$ J s (the data are approximate).

(a) 530.9 nm (b) about 6.7 × 10⁻³⁴ J s

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{256.7\times 10^{-9}}=7.74\times 10^{-19}$ J $=4.83$ eV. Maximum K.E. $=e{V}_0=0.35$ eV.

$W_0=4.83-0.35$

4.48 eV

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{150\times 10^{-12}}=1.325\times 10^{-15}$ J; K.E. $=\frac{1}{2}(9.109\times 10^{-31})(1.5\times 10^7{)}^2=1.025\times 10^{-16}$ J.

Binding energy $=1.325\times 10^{-15}-1.025\times 10^{-16}=1.22\times 10^{-15}$ J $=\frac{1.22\times 10^{-15}}{1.602\times 10^{-19}}$ eV.

About 7.6 × 10³ eV (1.22 × 10⁻¹⁵ J)

$\nu =\frac{3\times 10^8}{1285\times 10^{-9}}=2.335\times 10^{14}$ Hz. $\frac{1}{9}-\frac{1}{n^2}=\frac{2.335\times 10^{14}}{3.29\times 10^{15}}=0.0710\Rightarrow \frac{1}{n^2}=0.0401\Rightarrow n^2=25$.

n = 5; infrared region

$r_n=52.9n^2$ pm: $\frac{1322.5}{52.9}=25\Rightarrow n_i=5$; $\frac{211.6}{52.9}=4\Rightarrow n_f=2$.

$\Delta E=2.18\times 10^{-18}\left(\frac{1}{4}-\frac{1}{25}\right)=4.578\times 10^{-19}$ J; $\lambda =\frac{hc}{\Delta E}=434$ nm.

434 nm; Balmer series, visible region

$\lambda =\frac{6.626\times 10^{-34}}{9.109\times 10^{-31}\times 1.6\times 10^6}=4.55\times 10^{-10}$ m.

455 pm

$v=\frac{h}{m\lambda }=\frac{6.626\times 10^{-34}}{1.675\times 10^{-27}\times 800\times 10^{-12}}$

494.5 m s⁻¹

$\lambda =\frac{6.626\times 10^{-34}}{9.109\times 10^{-31}\times 2.19\times 10^6}=3.32\times 10^{-10}$ m (equal to $2\pi r_1$, as in 2.32).

332 pm

$\lambda =\frac{6.626\times 10^{-34}}{0.1\times 4.37\times 10^5}$: far too small to detect.

1.516 × 10⁻³⁸ m

$\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626\times 10^{-34}}{4\times 3.14\times 0.002\times 10^{-9}}=2.64\times 10^{-23}$ kg m s⁻¹.

The given momentum $=\frac{h}{4\pi \times 0.05\times 10^{-9}}=1.05\times 10^{-24}$ kg m s⁻¹, smaller than its own uncertainty, so it cannot be defined meaningfully.

Δp = 2.64 × 10⁻²³ kg m s⁻¹; yes, the value is smaller than the uncertainty.

Orbitals: 1 = 4d (n + l = 6), 2 = 3d (5), 3 = 4p (5), 4 = 3d (5), 5 = 3p (4), 6 = 4p (5). Lower n + l means lower energy; for equal n + l, the lower n comes first (3d before 4p).

5 < 2 = 4 < 3 = 6 < 1; 2 and 4 (3d), and 3 and 6 (4p), have the same energy.

The 4p electrons are farthest from the nucleus and most shielded by the inner electrons.

The 4p electrons

(i) 2s is closer to the nucleus. (ii) 4d penetrates more than 4f. (iii) 3p penetrates more than 3d.

(i) 2s (ii) 4d (iii) 3p

Si has one more proton (14 vs 13), and its extra electron is in the same shell, which shields poorly.

Silicon's 3p electrons

(a) P [Ne]3s² 3p³ (b) Si [Ne]3s² 3p² (c) Cr [Ar]3d⁵ 4s¹ (d) Fe [Ar]3d⁶ 4s² (e) Kr [Ar]3d¹⁰ 4s² 4p⁶ (Hund's rule).

(a) 3 (b) 2 (c) 6 (d) 4 (e) 0

(a) l = 0, 1, 2, 3: 4s, 4p, 4d, 4f. (b) n = 4 has 16 orbitals, each with one electron of mₛ = −½.

(a) 4 (b) 16