Atomic Structure

Number of protons, neutrons and electrons in $_6^{13}$C are

Atomic number = 6

Atomic number = number of protons = number of electrons = 6

$\therefore$ Protons (p) = 6

Electrons (e) = 6

Atomic mass = number of protons + number of neutrons

13 = 6 + x $\Rightarrow$ x = 7 (Atomic mass = 13)

Mass of proton $(p)=1.6726 \times 10^{-27} \mathrm{~kg}$

Mass of neutron $(n)=1.6749 \times 10^{-27} \mathrm{~kg}$

Mass of electron $(e)=9.1 \times 10^{-31} \mathrm{~kg}$

Mass of proton is 1836.16 times heavier than electron.

Mass of neutron is 1838.68 times heavier than electron.

Ratio of mass of electron, proton and neutron will be

Mass of $e:$ Mass of $p:$ Mass of $n$

$1: 1836.16: 1838.68$

Number of electrons present in the following species

Atomic number $=$ number of electrons in neutral species.

(i) $\mathrm{Be}^{2+} \Rightarrow$ Atomic number $=4$

It donate 2 electrons and acquire +2 charge.

$\therefore$ Number of electrons $=2$

(ii) $\mathrm{H}^{+} \Rightarrow$ Atomic number $=1$

It donate 1 electron and acquire +1 charge.

Number of electrons $=0$

(iii) $\mathrm{Na}^{+} \Rightarrow$ Atomic number $=11$

+1 charge represent donation of 1 electron.

Number of electrons $=10$

(iv) $\mathrm{Mg}^{+} \Rightarrow$ Atomic number $=12$

$\mathrm{Mg}^{+}$ represent donation of 1 electron.

Number of electrons $=11$

$\begin{aligned} \therefore \quad \mathrm{Be}^{2+} & =2, \mathrm{H}^{+}=0 \\ \mathrm{Na}^{+} & =10, \mathrm{Mg}^{+}=11 \end{aligned}$

Metals emit electrons when light shines upon them. The phenomenon is called as photoelectric effect and the electrons emitted in such manner are called photoelectrons.

Relation between kinetic energy and frequency is kinetic energy $=\phi-h \mathrm{v}$

Here $\phi=$ metal work function

Kinetic energy of photoelectrons depends on frequency of incident light and metal work function. $\mathrm{Cu}$ has maximum metal work function's value.

Hence, $\mathrm{Cu}$ have minimum kinetic energy of ejected photoelectrons than alkali metals.

We know that, energy $=h c / \lambda$

Here, $h=$ Planck's constant (Js )

$c=$ speed of light $(\mathrm{m} / \mathrm{s})$

$\lambda=$ wavelength $(\mathrm{m})$

Now for energy $E_1=25 \mathrm{eV}$

$E_1=h c / \lambda_1 \Rightarrow 25=h c / \lambda_1$ ..... (i)

Similarly for energy,

$E_2=h c / \lambda_2$

$100=h c / \lambda_2$ ..... (ii)

By divide $E_1$ and $E_2$, we get

$\begin{aligned} & 25 / 100=\lambda_2 / \lambda_1 \Rightarrow 1 / 4=\lambda_2 / \lambda_1 \\ & \lambda_1: \lambda_2=4: 1 \text { or } \lambda_1=4 \lambda_2 \end{aligned}$

Let us consider, particle $A$ having velocity $=v_A$ and particle $B$ having velocity $=v_B$

According to de-Broglie wavelength, $\lambda=\frac{h}{m v}$

here, $h=$ Planck constant

$m=$ mass of electron

$v=$ velocity of particle

Now, $\lambda_A=\frac{h}{m_A v_A}$ and $\lambda_B=\frac{h}{m_B v_B}$

Wavelength of $A$ is found to be double than of $B$.

$\text { i.e. } \lambda_A=2 \times \lambda_B$ ..... (i)

Put value of $\lambda_A$ and $\lambda_B$ in Eq. (i)

$\begin{array}{rlrl} 2 \times \frac{h}{m_B v_B} & =\frac{h}{m_A v_A} & & \left(\text { For } v_A=v_B\right) \\ \frac{2}{m_B} & =\frac{1}{m_A} & {\left[m_A=\frac{m_B}{2}\right]}\end{array}$

Here, mass of $A$ is half that of $B$.

Number of electrons in lithium ion (Li⁺) is 2.

Number of electrons in helium is 2.

The spectrum of an atom depend on the number of electrons present in it.

Li$^+$ (Z = 3)

He (Z = 2)

(Both have same electrons = 2)

Bohr radius of hydrogen atom is first orbit in H-atom.

Bohr radius $({r_n}) = \left( {{{{n^2}} \over Z}} \right)\left( {{{{n^2}} \over {4{\pi ^2}m{e^2}}}} \right)$

${r_n} = {{{n^2}} \over Z} \times 0.529\,\mathop A\limits^o $

For first orbit $(n = 1),{r_1} = \left( {{1 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ ..... (i)

For second orbit $(n = 2),{r_2} = \left( {{4 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ ..... (ii)

For third orbit $(n = 3),{r_3} = \left( {{9 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ .... (iii)

Now divided (iii) by (i), we get

${r_3} = 9{r_1}$

The radius of 3rd orbit is 9 times the radius of 1st orbit.

The magnetic moment is given by $\mu = \sqrt {n(n + 2)} $ BM. (where, n = number of unpaired electron).

Electronic configuration of

$T{i^{2 + }}(22):1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}$

$\because$ Number of unpaired electron = 2

$\therefore$ $\mu = \sqrt {2(2 + 2)} = \sqrt 8 $

Energy for H-like atom, $(E) \propto {{{Z^2}} \over {{n^2}}}$

For hydrogen, Z = 1, n = 1

For lithium Z = 3, n = ?

$\therefore$ For same energy,

${E_H} = {E_{Li}} \Rightarrow {{Z_H^2} \over {n_H^2}} = {{Z_{Li}^2} \over {n_{Li}^2}}$

$\therefore$ $n_{Li}^2 = {{Z_{Li}^2 \times n_H^2} \over {Z_H^2}} \Rightarrow n_{Li}^2 = {{{3^2} \times 1} \over 1}$

$n_{Li}^2 = 3$

While angular momentum (nh/2$\pi$) is independent of atomic number.

$\therefore$ For 3s-orbital of lithium atom, energy and angular momentum is equal to that of 1s orbital of hydrogen atom.

The energy required for ionisation

H-atom is given by, $E = {{ - kZ{e^2}} \over {2r}}$

where, Z = atomic number

e = charge on proton/electron

r = radius of that electron

$\therefore$ $E \propto {1 \over r}$ or E $\propto$ r^$-$1

$\therefore$ n = $-$1

Rutherford model could not explain.

(i) Electronic structure of an atom.

(ii) Stability of an atom.

The maximum number of possible electrons in a subshell with $n=3, l=2$ are 10 .

Because for, $n=3, l=2$, i.e. $d$-subshell has 10 electrons and 5 -orbital.

Hence, option (a) is the correct answer.

Given, uncertainty in position $(\Delta x)=1 \times 10^4 \mathrm{~m}$

uncertainty in velocity

$ (\Delta v)=0.627 \times 10^{-20} \mathrm{~ms}^{-1} $

Also,

$ \begin{aligned} & \because \quad \Delta x \cdot \Delta v=\frac{h}{4 \pi \cdot m} \\ & \text { } \begin{aligned} where, h & =\text { Plank's constant } \\ & =6.627 \times 10^{-34} \mathrm{Js} \\ m & =\text { to find (mass of particle) } \\ \therefore \quad m & =\frac{h}{\Delta x \cdot \Delta v \cdot 4 \pi} \\ & =\frac{6.627 \times 10^{-34}}{10^{-8} \times 6.627 \times 10^{-20} \times 4 \pi} \\ & =\frac{10^{-34} \times 10^{28}}{4 \pi} \text { in } \mathrm{kg} \\ \therefore \quad m & =\frac{10^{-34} \times 10^{28}}{4 \pi} \Rightarrow m=\frac{10^{-6}}{4 \pi} \mathrm{~kg} \end{aligned} \end{aligned} $

Hence, option (d) is the correct answer.

From de-Broglie's equation, $\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m \times \mathrm{KE}}}$

⇒ For an electrons, $\lambda \propto \frac{1}{\sqrt{\mathrm{KE}}} \Rightarrow \frac{\lambda_{16}}{\lambda_9}=\sqrt{\frac{9}{16}}=3: 4$

Regarding photoelectric effect, the statement of option-(c) is not correct. Because, the electrons are ejected from the metal surface as soon as the beam of light strikes to the surface.

This model could not explain the overall neutrality of the atom, because he proposed that the positive charge is spread over a sphere in which the electrons are embedded to make the atom as a whole neutral, but it was not consistant with the results. Hence, discarded.

In presence of electric and magnetic field, the cathode rays travel in straight line is not the result. Actually, they are deflected by electric and magnetic field.

When the rays are passed between two electrically charged plates, these are deflected towards positively charged plate.

In magnetic field, they are deflected at a right angle to both the field and to the direction of particle.

Given, $E=\frac{-R_{\mathrm{H}}}{16}$

We know that, $E=\frac{-R_{\mathrm{H}}}{n^2}=\frac{-R_{\mathrm{H}}}{16}, \therefore n=4$

Thus, degeneracy for $n=4$ will be,

Given, kinetic energy of $\mathrm{H}^{+}$ion $=1.68 \mathrm{eV}$

Mass of proton $=1.6726 \times 10^{-27} \mathrm{~kg}$

We calculate, the wavelength of $\mathrm{H}^{+}$ion as follows:

$ \lambda=\frac{h}{\sqrt{2 E_m}} $

Here, $E=$ kinetic energy, $m=$ mass of proton

$ \begin{aligned} h & =6.62 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} . \\ & =\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times E}} \\ & =\frac{0.286}{\sqrt{E(\mathrm{eV})}} \mathop {\rm{A}}\limits^{\rm{o}} =\frac{0.286}{\sqrt{1.68}}\mathop {\rm{A}}\limits^{\rm{o}} =0.220\mathop {\rm{A}}\limits^{\rm{o}} =0.022 \mathrm{~nm} \end{aligned} $

As we know,

$\nu = {1 \over \lambda } = {{{{(IE)}_H}{Z^2}} \over {hc}}x - \left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$

Given, n₁ = 2, n₂ = 5

${(IE)_H} = 13.6\,eV = 13.6 \times 1.6 \times {10^{ - 19}}\,J$

$\nu = {1 \over \lambda } = {{13.6 \times 1.6 \times {{10}^{ - 19}}\,J} \over {6.6 \times {{10}^{ - 34}}\,Js \times 3 \times {{10}^8}\,m{s^{ - 1}}}}\left[ {{1 \over {{2^2}}} - {1 \over {{5^2}}}} \right]$

$ = 2.2988 \times {10^6}\,{m^{ - 1}}$

$\therefore$ $\lambda = {1 \over \nu } \Rightarrow \lambda = 4.35 \times {10^{ - 7}}\,m = 435 \times {10^{ - 9}}\,m$

= 435 nm

If n = n, then I = 0, 1, 2 .... n

Thus, energy order is,

1s < 1p < 2s < 2p < 3s < 2d < 3p

Sc (21) will have electronic configuration as :

1s² 1p⁶ 2s² 2p⁶ 3s² 2d³