(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 $

(b)

${NH}_3 {A}=179$

179 of ${NH}_3 6.023 \times 10^{23} {NH}_3$ molecules

34 {mg } of ${NH}_3 \frac{6.023 \times 10^{23}}{17 \times 10^3} \times 34=12.046 \times 10^{20} {NH}_3$ molecules

${NH}_3 \Rightarrow 10 { protons } \Rightarrow 34 { mg} { NH_3} \Rightarrow 1.2046 \times 10^{22} \text { protons }$

Mass of proton =$1.2046 \times 10^{22} \times 1.672 \times 10^{-27} {kg}$ = 20.1 mg

$_z^AX$

How many neutrons and protons are there in ${ }_{26}^{56} \mathrm{Fe}$ ?

An ion with mass number 37 possesses one unit of negative charge. If the ion contains $11.1$percent more neutrons than the electrons, find the symbol of the ion.

(a)

${ }_{26}^{56} \mathrm{Fe} .$

$Z = 26 = \text{number of protons}$

$A = \text { number of protons } + \text {number of neutrons }$

$\text { Number of neutrons }30$

(b)

$A = 37 \quad N_e = N_p+1$

$N_n=1.111 \mathrm { ~ N_e}$

$N_p+N_M=37$

$N_e-1+1.111 ~ N_e=37$

$2.111 ~ N_e=38$

$N_e=\frac{38}{2.111}=18$

$N_p=17 \Rightarrow Z=17$

$N_n=A-Z=20$

${}_{17}^{37} Cl^-$

Find energy of each of the photons which

(i) Correspond to light of frequency $3 \times 10^{15} \mathrm{Hz}$.

(ii) Have wavelength of $0.50 \mathring{A}$.

Find the wavelength of the photon whose period is $2.0 \times 10^{-10} \mathrm{~s}$.

$E=.h \nu=\frac{h c}{\lambda}=h c \bar{\nu}=\frac{h}{\tau}$

$\nu=\frac{c}{\lambda}=c \bar{\nu}=\frac{1}{\tau}$

(a) - (i)

$\nu=3 \times 10^{15} \mathrm{ ~ Hz} \Rightarrow E=h \nu$

$=6.626 \times 10^{-34} {J} { s} 3\times 10^{15}{s}^{+}$

$=19.88 \times 10^{-19} \mathrm{ ~ J}$

(ii)

$\lambda=0.5 \mathring{A}=5 \times 10^{-11} { m}$

$E=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8 .{Js} {ms^+}}{5 \times 10^{-11} { m}}=3.976 \times 10^{-15} \mathrm{ ~ J}$

(b)

$\tau=2 \times 10^{-10} \mathrm{ ~ s} \Rightarrow \nu=\frac{1}{\tau}=0.5 \times 10^{10} \mathrm{ ~ s}^{-1}$

$\nu=\frac{c}{\lambda} \Rightarrow \lambda=\frac{c}{\nu}=\frac{3 \times 10^8 \mathrm{ ~ ms}^{-1}}{0.5 \times 10^{10} \mathrm{ ~ s}^{-1}}=0.06 \mathrm{ ~ m}$

$\bar \nu=\frac{1}{\lambda}$

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 (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron. $(\left.1 \mathrm{eV}=1.6020 \times 10^{-19} \mathrm{~ J}\right)$.

$\lambda=4 \times 10^{-7} \mathrm{~ m}$

$\phi_0=2.13 \mathrm{~ eV}$

(i) $E=\frac{h c}{\lambda} =\frac{6.626 \times 10^{-34} \times 3 \times 10^8 \mathrm{Js ~ ms}^{-1}}{4 \times 10^{-7} \mathrm{~ m}} =4.07 \times 10^{-19} \mathrm{~ J}$

• = $\frac{4.07 \times 10^{-19}}{1.602 \times 10^{-19}} {eV}=3.10{eV}$

(ii) $e_{K E}=E-\phi_0=0.97 \mathrm{~ eV}$

(iii) $\frac{1}{2} m v^2=0.97 \times 1.602 \times 10^{-19} \mathrm{~ J}$

$v^2=\frac{2 \times 0.97 \times 1.602 \times 10^{-19}}{9.11 \times 10^{-31}} \mathrm{\frac{J}{Kg}} m^2 s^{-2}$

$v=5.84 \times 10^6 \mathrm{~ ms}^{-1}$

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 $n=2$ ?

How much energy is required to ionise a $\mathrm{H}$ atom if the electron occupies $n=4$ orbit?

$n = 4$

$n = 3$

$n = 2$

$n = 1$

$E_n=-2.18 \times 10^{-18}\left(\frac{Z^2}{n^2}\right) \mathrm{J}$

(a)

$E_4=-2.18 \times 10^{-18} \frac{1}{4^2} \mathrm{ J}$

$E_2=-2.18 \times 10^{-18} \frac{1}{2^2} \mathrm{~ J}$

$E_2-E_4=-2.18 \times 10^{-18}\left(\frac{1}{4}-\frac{1}{16}\right) \mathrm{J} =-4.087 \times 10^{-19} \mathrm{J }$

$\lambda=E=\frac{h c}{\lambda} \rightarrow \lambda = \frac{hc}{E}$

$=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.087 \times 10^{-19} \mathrm{J}} Jm =486.3 \mathrm{ nm}$

(b)

$E_f=0$

$E_i=E_4 \quad \Rightarrow \quad IE =\frac{2.18}{16} \times 10^{-18} \mathrm{ J} = 1.36 \times 10^{-19} \mathrm{ J}$

Calculate the energy required for the following process

$\mathrm {He}^+ (\mathrm {g}) \rightarrow \mathrm {H}e^{2+}(\mathrm {g}) + \mathrm {e}^-$

$E_n=-2.18 \times 10^{-18}\left(\frac{Z^2}{n^2}\right) \mathrm{J}$

$\mathrm{E} = -2.18\times 10 ^{-18} \times 4 \mathrm {J}$

$= \underbrace{-8.72 \times 10 ^{-18}} \mathrm {J}$

The mass of an electron is $9.1\times 10^{-31}\mathrm { kg.}$ If its $\mathrm {K.E.}$ is $3.0 \times 10 ^{-25} \mathrm { J}$, calculate its wavelength.

$\lambda = \frac{h}{p} = \frac{h}{mv}$

${K E}= \frac{p^2}{2 m}=3 \times 10^{-25} \mathrm{ J}$

$p^2=2 \times 9.11 \times 10^{-31} \times 3 \times 10^{-25} \mathrm { J} \mathrm {~ kg}$

${p}=7.39 \times 10^{-28} \mathrm{~ kg} \mathrm{~ m} \mathrm{~ s}^{-1}$

$\lambda=\frac{h}{p} =\frac{6.626 \times 10^{-34} \mathrm{Js}}{7.39 \times 10^{-28} {kg} { m} \mathrm{~ s}^{-1}} = 0.897 \times 10^{-6} {m} = 897 {nm}$

How many subshells are associated with $n=4?$

How many electrons will be present in the subshells having $m_s$ value of $-1 / 2$ for $n=4 ?$

$n=1,2,3, \cdots$

$l=0,1,2, \cdots, n-1$

$m_l=-l,-l+1, \cdots, l-1, l$

$m_s= \pm 1 / 2$

(a)

$\mathrm{n} = 4$

$l = 0, 1, 2, 3 \rightarrow 4 \text { subshells }$

$l=0 \rightarrow m_l = 0 \quad 1L \text{ Orbital } 4 s$

$1L = +\frac{1}{2},$ $\quad \qquad 1L = -\frac{1}{2}$

$l= 1 \rightarrow m_l = -1, 0, 1 (1L) (1L) (1L)$

$l= 2 \rightarrow m_l =\pm 2, \pm 1,0$

$l= 3 \rightarrow m_l =\pm 3, \pm 2, \pm 1,0$

$16 \text{ orbitals} (n^2)$

$\text{Number of} ~ e^- \rightarrow 32 (2n^2)$

(b) $\qquad 16 ~ e^-$

An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

$\underline{\text{ Increasing order of} ~n +l :}$

$1 s$ $2s$ $2 p$ $3 s$ $3 p$ $4 s$ $3 d$ $4 p$ $5 s \ldots$.

$\text { Number of } p=29=z \quad \mathrm{Cu}$

${ }_{29}^{64} \mathrm{Cu} \quad 1 s^2 \quad {2 s^2} \quad {2 p^6} \quad 3 s^2 \quad3 p^6 \quad 4 s \quad 3d$

$\underbrace{4s^2 \quad 3d^9}$

$4s^1 \quad 3d^{10}$

Arrange the following orbitals in increasing order of radial nodes, angular nodes, and total nodes:

1s, 2s, 2p, 3s, 3p, 3d

number of radial nodes $=n-1-1$

number of angular nodes $=1$

number of nodes $=n-1$

Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

(i) $1 \mathrm{~ s}$ and $2 \mathrm{~ s}$,

(ii) 4d and 4f,

(iii) 3d and 3p.

$1 s>2 s$

$4 d>4 f$

$3 p>3 d$