chemistry-class-11-unit-02-chapter-08-structure-of-the-atom-l-8-8-57jiayehoue

Structure Of The Atom L-8

Charge and mass of subatomic particles

Name	Symbol	Absolute charge/C	Relative charge	Mass/kg	Mass/u	Approx mass/u
Electron	$e$	$-1.602176 \times 10^{-19}$	$-1$	$9.109382 \times 10^{-31}$	$0.00054$	$0$
Proton	$p$	$+1.602176 \times 10^{-19}$	$+1$	$1.6726216 \times 10^{-27}$	$1.00727$	$1$
Neutron	$n$	$0$	$0$	$1.674927 \times 10^{-27}$	$1.00867$	$1$

Calculate the mass and charge of one mole of electrons
Find (a) the total number and (b) the total mass of protons in $34 \mathrm{mg}$ of $\mathrm{NH}_3$ at STP

Structure Of The Atom L-8

Charge and mass of subatomic particles

Question 1 - Calculate the mass and charge of one mole of electrons Find
- (a) The total number and
- (b) The total mass of protons in 34 mg of $NH_3$ at STP
Answer (a)
- $M=6.023 \times 10^{23} \times 9.11 \times 10^{-31} {kg }$ $=5.48 \times 10^{-7} { kg }$
- $Q=6.023 \times 10^{23} \times (-) {1.602 \times 10^{-19}} {C}$ $=96485 ~ C = 1 F$

Structure Of The Atom L-8

Charge and mass of subatomic particles

Answer (b)
${NH}_3 \quad, {A}=179$
$179 \text{ of } {NH}_3 \quad 6.023 \times 10^{23} \quad {NH}_3 \text { molecules }$
$34 {mg } \text { of }{NH}_3 \quad \frac{6.023 \times 10^{23}}{17 \times 10^3} \times 34=12.046 \times 10^{20} { ~NH}_3 \text { molecules }$
${NH}_3 \Rightarrow 10 \text { protons } \Rightarrow 34 {mg} {NH_3} \Rightarrow 1.2046 \times 10^{22} \text { protons }$
$\text { Mass of proton }=1.2046 \times 10^{22} \times 1.672 \times 10^{-27} {kg}$
$\text { Mass of proton } =20.1{ mg}$

Structure Of The Atom L-8

Atomoic mass and atomic number

Question 2
(a) How many neutrons and protons are there in ${ }_{26}^{56} \mathrm{Fe}$ ?
(b) An ion with mass number 37 possesses one unit of negative charge. If the ion contains $11.1$ % more neutrons than the electrons , find the symbol of the ion
Answer a
- ${ }_{26}^{56} \mathrm{Fe}$
- $Z = 26 = \text{number of protons}$
- $\text { Number of neutrons }30$
- $A = \text { number of protons } + \text {number of neutrons }$

Structure Of The Atom L-8

Atomoic mass and atomic number

Answer b
- A = 37
- $N_e = N_p+1$
- $N_n=1.111 { N_e}$
- $N_p+N_n=37$
- $N_e-1+1.111 ~ N_e=37$
- $2.111 ~ N_e=38$
- $N_e=\frac{38}{2.111}=18$
- $N_p=17 \Rightarrow Z=17$
- $N_n=A-Z=20$
- ${}_{17}^{37} Cl^-$

Structure Of The Atom L-8

Wavelength - wavenumber-frequency-time period-energy

Question 3 - Find energy of each of the photons which
(a) Correspond to light of frequency $3 \times 10^{15} {Hz}$
(b) Have wavelength of $0.50 \mathring {\mathrm A}$
(c) Find the wavelength of the photon whose period is $2.0 \times 10^{-10} \mathrm{~s}$
$E=h \nu=\frac{h c}{\lambda}=h c \bar{\nu}=\frac{h}{\tau}$
$\nu=\frac{c}{\lambda}=c \bar{\nu}=\frac{1}{\tau}$

Structure Of The Atom L-8

Wavelength - wavenumber-frequency-time period-energy

Answer (a)
- $\nu=3 \times 10^{15} \mathrm{ ~ Hz} \Rightarrow E=h \nu$
- $\nu=6.626 \times 10^{-34} { J} { s} \quad 3 \times 10^{15} { s}^{+}$
- $\nu=19.88 \times 10^{-19} { J}$

Structure Of The Atom L-8

Wavelength - wavenumber-frequency-time period-energy

Answer (b)
- $\lambda=0.5 \mathring {\mathrm A}=5 \times 10^{-11} { m}$
- $E=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8{Js} {ms^{-1}}}{5 \times 10^{-11} { m}}$
- $E=3.976 \times 10^{-15} { J}$

Structure Of The Atom L-8

Wavelength - wavenumber-frequency-time period-energy

Answer (c)
$\tau=2 \times 10^{-10} { s} \Rightarrow \nu=\frac{1}{\tau}=0.5 \times 10^{10} { s}^{-1}$
$\nu=\frac{c}{\lambda} \Rightarrow \lambda=\frac{c}{\nu}=\frac{3 \times 10^8 { ms}^{-1}}{0.5 \times 10^{10} { s}^{-1}}=0.06 { m}$
$\bar \nu=\frac{1}{\lambda}$

Structure Of The Atom L-8

Photoelectric effect

Question 4 - A photon of wavelength $4 \times 10^{-7} \mathrm{~ m}$ strikes on metal surface , the work function of the metal being $2.13 \mathrm{eV}$ . Calculate
(a) The energy of the photon (eV) ,
(b) The kinetic energy of the emission and
(c) The velocity of the photoelectron $(\left.1 \mathrm{eV}=1.6020 \times 10^{-19} \mathrm{~ J}\right)$

Structure Of The Atom L-8

Photoelectric effect

Answer (a)
- $\lambda=4 \times 10^{-7} \mathrm{~ m}$
- $\phi_0=2.13 \mathrm{~ eV}$
- $E=\frac{h c}{\lambda} =\frac{6.626 \times 10^{-34} \times 3 \times 10^8 {Js ~ ms}^{-1}}{4 \times 10^{-7} { m}}$
- $E=4.07 \times 10^{-19} {J}=\frac{4.07 \times 10^{-19}}{1.602 \times 10^{-19}} {eV}=3.10 {eV}$

Structure Of The Atom L-8

Photoelectric effect

Answer (b)
- $e_{K E}=E-\phi_0=0.97 { eV}$

Structure Of The Atom L-8

Photoelectric effect

Answer (c)
- $\frac{1}{2} m v^2=0.97 \times 1.602 \times 10^{-19} { J}$
- $v^2=\frac{2 \times 0.97 \times 1.602 \times 10^{-19}}{9.11 \times 10^{-31}} {\frac{J}{Kg}} m^2 s^{-2}$
- $v=5.84 \times 10^6 { ms}^{-1}$

Structure Of The Atom L-8

Hydrogen atom energy levels

Question 5
(a) What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2 ?
(b) How much energy is required to ionise a $\mathrm{H}$ atom if the electron occupies n = 4 orbit ?
- $n = 4$
- $n = 3$
- $n = 2$
- $n = 1$

Structure Of The Atom L-8

Hydrogen atom energy levels

Answer (a)
- $E_4=-2.18 \times 10^{-18} \frac{1}{4^2} { J}$
- $E_2=-2.18 \times 10^{-18} \frac{1}{2^2} { J}$
- $E_2-E_4=-2.18 \times 10^{-18}\left(\frac{1}{4}-\frac{1}{16}\right) {J}$
- $E_2-E_4=-4.087 \times 10^{-19} {J }$
- $\lambda=E=\frac{h c}{\lambda} \rightarrow \lambda = \frac{hc}{E}$
- $\lambda=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.087 \times 10^{-19} {J}} Jm =486.3 { nm}$

Structure Of The Atom L-8

Hydrogen atom energy levels

Answer (b)
- $E_f=0$
- $E_i=E_4 \quad \Rightarrow \quad IE =\frac{2.18}{16} \times 10^{-18} { J}$
- $IE=1.36 \times 10^{-19} { J}$

Structure Of The Atom L-8

Hydrogen-like systems

Question 6 - Calculate the energy required for the following process ${He}^+ ({g}) \rightarrow {H}e^{2+}({g}) + {e}^-$
Answer
- $E_n=-2.18 \times 10^{-18}\left(\frac{Z^2}{n^2}\right) {J}$
- ${E} = -2.18\times 10 ^{-18} \times 4 { J}$
- $E= \underbrace{-8.72 \times 10 ^{-18}} { J}$

Structure Of The Atom L-8

De-brogile's hypothesis

Question 7 - The mass of an electron is $9.1\times 10^{-31}{ kg.}$ If its ${K.E}$ is $3.0 \times 10 ^{-25} { J}$ , calculate its wavelength $\lambda = \frac{h}{p} = \frac{h}{mv}$
Answer
- ${K E}= \frac{p^2}{2 m}=3 \times 10^{-25} { J}$
- $p^2=2 \times 9.11 \times 10^{-31} \times 3 \times 10^{-25} { J} { kg}$
- ${p}=7.39 \times 10^{-28} { kg} { m} { s}^{-1}$
- $\lambda=\frac{h}{p} =\frac{6.626 \times 10^{-34} {Js}}{7.39 \times 10^{-28} { kg} { m} {s}^{-1}}$
- $\lambda=0.897 \times 10^{-6} { m}$
- $\lambda=897 {nm}$

Structure Of The Atom L-8

Quantum numbers

Question 8
(a) How many subshells are associated with n=4 ?
(b) How many electrons will be present in the subshells having $m_s$ value of $-1 / 2$ for $n=4 ?$
$n=1,2,3, \cdots$
$l=0,1,2, \cdots, n-1$
$m_l=-l,-l+1, \cdots, l-1, l$
$m_s= \pm 1 / 2$

Structure Of The Atom L-8

Quantum numbers

Answer (a)
- ${n} = 4$
- $l = 0, 1, 2, 3 \rightarrow 4 \text { subshells }$
- $l=0 \rightarrow m_l = 0 \quad 1L \text{ orbital } 4 s$
- $1L = +\frac{1}{2}$
- $1L = -\frac{1}{2}$
- $l= 1 \rightarrow m_l = -1, 0, 1 (1L) (1L) (1L)$
- $l= 2 \rightarrow m_l =\pm 2, \pm 1,0$
- $l= 3 \rightarrow m_l =\pm 3, \pm 2, \pm 1,0$
- $16 \text{ orbitals} (n^2)$
- $\text{number of} ~ e^- \rightarrow 32 (2n^2)$

Structure Of The Atom L-8

Quantum numbers

Answer (b)
- $16 ~ e^-$

Structure Of The Atom L-8

Electronic configuration

Question 9 - An atom of an element contains 29 electrons and 35 neutrons. Deduce
(a) The number of protons and
(b) The electronic configuration of the element
Answer
- $\underline{\text{ Increasing order of} ~n +l :}$
- $1 s$ $2s$ $2 p$ $3 s$ $3 p$ $4 s$ $3 d$ $4 p$ $5 s \ldots$
- $\text { Number of } p=29=z \quad {Cu}$
- ${ }_{29}^{64} {Cu} \quad 1 s^2 \quad {2 s^2} \quad {2 p^6} \quad 3 s^2 \quad3 p^6 \quad 4 s \quad 3d$
- $\underbrace{4s^2 \quad 3d^9}$
- $4s^1 \quad 3d^{10}$

Structure Of The Atom L-8

Nodes in orbitals

Question 10 - Arrange the following orbitals in increasing order of radial nodes , angular nodes, and total nodes - 1s, 2s, 2p, 3s, 3p, 3d
$\text {Number of radial nodes} =n-1-1$
$\text {Number of angular nodes}=1$
$\text {Number of nodes}=n-1$
Answer

Nodes	2s	2p	3s	3p	3d
$\text {Number of angular nodes}$	0	1	0	1	2
$\text {Number of radial nodes}$	1	0	2	1	0
$\text {Number of total nodes}$	1	1	2	2	2

Structure Of The Atom L-8

Effective nuclear charge

Question 11 - Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge ?
(a) 1s and 2s $\quad$
(b) 4d and 4f $\quad$
(c) 3d and 3p
Answer
(a) $1{s} > 2{s}$
(b) $4{d} > 4{f}$
(c) $3{p} > 3{d}$

Structure Of The Atom L-8

Structure of atom