Chapter 2: Structure of Atom

Ans: 

Total number of electrons in an atom for a value of n = $2{{n}^{2}}$ 

$\therefore $ For n = 4,

Total number of electrons = $2{{(4)}^{2}}=32$

The given element has a fully filled orbital as $1{{s}^{2}}2{{s}^{2}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}$ 

Hence, all the electrons are paired. 

$\therefore $ Number of electrons (having n = 4 and ${{m}_{s}}=-\frac{1}{2}$ ) = 16 

Ans: n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

Ans: Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

$mvr=n\frac{h}{2\pi }$ ----------- (i)

Where, n = 1, 2, 3 …

According to de Broglie’s equation:

 $\lambda =\frac{h}{mv}$ Or $mv=\frac{h}{\lambda }$ ------- (ii)

Substituting the value of ‘mv’ from expression (ii) in expression (i):

$\frac{hr}{\lambda }=n\frac{h}{2\pi }$ ----------------- (iii)

Since $2\pi r$  represents the circumference of the Bohr orbit (r), it is proved by equation (iii) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Ans: for $H{{e}^{+}}$ion, the wave number $\overset{-}{\mathop{v}}\,$ associated with the Balmer transition, n=4 to n =2 given by:

 \[\overset{-}{\mathop{v}}\,=\frac{1}{\lambda }=R{{Z}^{2}}\left( \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}^{2}}_{2}} \right)\] 

Where, ${{n}_{1}}=2and{{n}_{2}}=4$ , Z = atomic number of Helium

  \[\overset{-}{\mathop{v}}\,=\frac{1}{\lambda }=R{{2}^{2}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{4}\] 

 \[\Rightarrow \lambda =\frac{4}{3R}\] 

According to the question, the desired transition for hydrogen will have the same wavelength as that of $H{{e}^{+}}$.

\[R{{(1)}^{2}}\left( \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right)=\frac{3R}{4}\] 

 \[\left( \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right)=\frac{3}{4}\] -------- (1)

By hit and trial method, the equality given by equation (1) is true only when ${{n}_{1}}=1$ and ${{n}_{2}}=2$ 

The transition for ${{n}_{2}}=2$to ${{n}_{1}}=1$ in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of $H{{e}^{+}}$spectrum.

Ans: Energy associated with hydrogen-like species is given by,

\[{{E}_{n}}=\frac{-(2.18\times {{10}^{-18}}){{Z}^{2}}}{{{n}^{2}}}\]

For ground state of hydrogen atom,

 \[\Delta E={{E}_{\infty }}-{{E}_{1}}\] 

\[\Rightarrow \Delta E=0-\left[ -2.18\times {{10}^{-18}}\left( \frac{{{(1)}^{2}}}{{{(1)}^{2}}} \right) \right]J\] 

\[\therefore \Delta E=2.18\times {{10}^{-28}}J\] 

For the given process, $H{{e}^{+}}_{(g)}\to H{{e}_{(g)}}^{2+}+{{e}^{-}}$

An electron is removed from n = 1 to n = ∞.

\[\Delta E={{E}_{\infty }}-{{E}_{1}}\]

\[\Rightarrow \Delta E=0-\left[ -2.18\times {{10}^{-18}}\left( \frac{{{(2)}^{2}}}{{{(1)}^{2}}} \right) \right]J\] 

 \[\therefore \Delta E=8.72\times {{10}^{-18}}J\] 

The energy required for the process $8.72\times {{10}^{-18}}J$ 

Ans: 1 m = 100 cm

1cm = ${{10}^{-2}}m$ 

Length of the scale = 20 cm = $20\times {{10}^{-2}}m$ 

Diameter of a carbon atom = 0.15 nm = $0.15\times {{10}^{-9}}m$ 

One carbon atom occupies $0.15\times {{10}^{-9}}m$

Number of carbon atoms that can be placed in a straight line = \[\frac{20\times {{10}^{-2}}m}{0.15\times {{10}^{-9}}m}=1.33\times {{10}^{9}}\] 

Ans: Length of the given arrangement = 2.4 cm

Number of carbon atoms present = $2\times {{10}^{8}}$

$\therefore $ Diameter of carbon atom = $=\frac{2.4\times {{10}^{-2}}}{2\times {{10}^{8}}}=1.2\times {{10}^{-16}}m$ 

Radius of carbon atom = $\frac{diameter}{2}=\frac{1.2\times {{10}^{-10}}m}{2}=6\times {{10}^{-11}}m$ 

Ans: Radius of zinc atom = $\frac{diameter}{2}=\frac{2.6{{A}^{0}}}{2}=1.3\times {{10}^{-11}}m=13\times {{10}^{-12}}m=13pm$ 

Ans: Length of the arrangement = 1.6 cm = $1.6\times {{10}^{-2}}m$ 

Diameter of zinc atom = $2.6{{A}^{0}}$= $2.6\times {{10}^{-10}}m$ 

Number of zinc atoms present in the arrangement =

\[\frac{1.6\times {{10}^{-2}}m}{2.6\times {{10}^{-10}}m}=0.6153\times {{10}^{8}}m\] 

Ans: Charge on one electron = $1.6022\times {{10}^{-19}}C$

\[\Rightarrow 1.6022\times {{10}^{-19}}C\] charge is carried by 1 electron.

Number of electrons carrying a charge of $2.5\times {{10}^{-16}}C$

\[=1\left( \frac{2.5\times {{10}^{-16}}C}{1.6022\times {{10}^{-19}}} \right)=1.560\times {{10}^{3}}C=1560C\] 

Ans: Charge on the oil drop = $-1.282\times {{10}^{-18}}C$

Charge on one electron = $1.6022\times {{10}^{-19}}C$

$\therefore $ Number of electrons present on the oil drop = $=\frac{1.282\times {{10}^{-18}}C}{1.6022\times {{10}^{-19}}C}=0.8001\times {{10}^{1}}=8.0$ 

Ans: A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms. 

Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of $\alpha $-particles (positively charged).