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NCERT Exemplar · Class 9 Science Structure of the Atom

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Long Answer Questions 36–43 (part 5 of 5)

  1. Exercise 36

    Why do Helium, Neon and Argon have a zero valency?

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    NCERT’s answer
    Helium has two electrons in its only energy shell, while Argon and Neon have $\displaystyle 8$ electrons in their valence shells. As these have maximum number of electrons in their valence shells, they do not have any tendency to combine with other elements. Hence, they have a valency equal to zero.
    \[\text{He: } K=2 \qquad \text{Ne: } K=2,\ L=8 \qquad \text{Ar: } K=2,\ L=8,\ M=8 \]Each outermost shell is already full — duplet for He, octet for Ne and Ar.\[\text{octet/duplet complete} \Rightarrow \text{no electron gained, lost or shared} \]Valency is the number of electrons gained, lost or shared to complete the octet/duplet, so\[\text{Valency(He)} = \text{Valency(Ne)} = \text{Valency(Ar)} = 0 \]NCERT_Solution_Class9_Science_Exemplar_Ch4_Q36Answer: Zero — their outermost shells already hold the full duplet/octet, so no electron needs to be gained, lost or shared.

  2. Exercise 37

    The ratio of the radii of hydrogen atom and its nucleus is ~ 105\displaystyle 10^{5}. Assuming the atom and the nucleus to be spherical, (i) what will be the ratio of their sizes? (ii) If atom is represented by planet earth ‘Re\displaystyle R_{e}’ = 6.4\displaystyle 6.4 ×106\displaystyle 10^{6} m, estimate the size of the nucleus.

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    (i)
    Volume of the sphere = r $\displaystyle 3$ π Let R be the radius of the atom and r be that of the nucleus. ⇒ R = \(\displaystyle 10^{5}\) r Volume of the atom = R $\displaystyle 3$ π = $\displaystyle 4$ $\displaystyle 3$ π (\(\displaystyle 10^{5}\)r)$\displaystyle 3$ (Q R = \(\displaystyle 10^{5}\)r ) = r $\displaystyle 3$ π × Volume of the nucleus = r $\displaystyle 3$π Ratio of the size of atom to that of nucleus = $\displaystyle 4$ × $\displaystyle 10$ r = $\displaystyle 10$ r π π × (ii) If the atom is represented by the planet earth (\(\displaystyle R_{e}\)= $\displaystyle 6.4$×\(\displaystyle 10^{6}\)m) then the radius of the nucleus would be e n R r = n $\displaystyle 6.4$ ×$\displaystyle 10$ m r = = $\displaystyle 6.4$×$\displaystyle 10$ m = $\displaystyle 64$ m.
    \[\frac{r_{\text{atom}}}{r_{\text{nucleus}}} = 10^{5} \]Volume of a sphere scales as \(\displaystyle r^{3}\), so\[\frac{V_{\text{atom}}}{V_{\text{nucleus}}} = \left(\frac{r_{\text{atom}}}{r_{\text{nucleus}}}\right)^{3} = (10^{5})^{3} = 10^{15} \]Taking the atom's radius as Earth's radius:\[R_{e} = 6.4\times10^{6}\ \text{m} \] \[r_{\text{nucleus}} = \frac{R_{e}}{10^{5}} = \frac{6.4\times10^{6}}{10^{5}}\ \text{m} = 64\ \text{m} \]Answer: (i) Volume ratio \(\displaystyle =10^{15}\). (ii) Nucleus radius \(\displaystyle \approx 64\ \text{m}\).

  3. Exercise 38

    Enlist the conclusions drawn by Rutherford from his α-ray scattering experiment.

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    NCERT’s answer
    Rutherford concluded from the α-particle scattering experiment that- (i) Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected. (ii) Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space. (iii) A very small fraction of α-particles were deflected by \(\displaystyle 180^{0}\), indicating that all the positive charges and mass of the gold atom were concentrated in a very small volume within the atom. From the data he also calculated that the radius of the nucleus is about \(\displaystyle 10^{5}\) times less than the radius of the atom.
    \(\displaystyle \alpha\)-particles fired at thin gold foil:Most passed straight through undeviated \(\displaystyle \Rightarrow\) the atom is mostly empty space.A few were deflected through small angles \(\displaystyle \Rightarrow\) a positively charged region repels them.A very small fraction (about $\displaystyle 1$ in $\displaystyle 12000$) bounced back through nearly \(\displaystyle 180^{\circ}\) \(\displaystyle \Rightarrow\) that positive charge, with nearly the whole mass, is concentrated in one tiny central volume — the nucleus.\[r_{\text{nucleus}} \sim 10^{-15}\ \text{m} \ll r_{\text{atom}} \sim 10^{-10}\ \text{m} \]Answer: The atom is mostly empty space, with a tiny, dense, positively charged nucleus at its centre carrying almost all the mass.

  4. Exercise 39

    In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?

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    NCERT’s answer
    Rutherford proposed a model in which electrons revolve around the nucleus in well-defined orbits. There is a positively charged centre in an atom called the nucleus. He also proposed that the size of the nucleus is very small as compared to the size of the atom and nearly all the mass of an atom is centred in the nucleus. Whereas, Thomson proposed the model of an atom to be similar to a christmas pudding. The electrons are studded like currants in a positively charged sphere like christmas pudding and the mass of the atom was supposed to be uniformly distributed.
    Thomson's model: positive charge spread uniformly through the whole sphere of the atom, with electrons studded inside it — like seeds in a watermelon; mass and charge both fill the entire volume.\[\text{Rutherford's model} \Rightarrow \text{positive charge \& mass concentrated in a tiny central nucleus} \]Electrons revolve around this nucleus in circular orbits, and most of the atom's volume is empty space — unlike Thomson's uniformly filled sphere.Answer: Thomson spread the positive charge through the whole atom; Rutherford concentrated it, with almost all the mass, in a tiny central nucleus, with electrons orbiting through otherwise empty space.

  5. Exercise 40

    What were the drawbacks of Rutherford’s model of an atom?

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    NCERT’s answer
    The orbital revolution of the electron is not expected to be stable. Any particle in a circular orbit would undergo a acceleration and the charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know.
    Stability: classical electromagnetic theory says an accelerating charge radiates energy continuously.\[\text{electron revolves (accelerates)} \Rightarrow \text{radiates energy continuously} \Rightarrow \text{energy}\downarrow \Rightarrow \text{orbit radius}\downarrow \Rightarrow \text{electron spirals into nucleus} \]This predicts an unstable atom, contradicting the observed stability of matter.Arrangement: the model said nothing about how electrons are arranged or distributed among orbits around the nucleus.Answer: It could not explain why atoms are stable -- by classical theory the orbiting electron should spiral into the nucleus -- and it said nothing about how electrons are arranged.

  6. Exercise 41

    What are the postulates of Bohr’s model of an atom?

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    NCERT’s answer
    The postulates put forth by Neils Bohr’s about the model of an atom: (i) Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom. (ii) While revolving in discrete orbits the electrons do not radiate energy. These orbits are called energy levels. Energy levels in an atom are shown by circles. These orbits are represented by the letters K,L,M,N,… or the numbers, n=$\displaystyle 1,2,3,4$,….
    1. Electrons revolve only in certain fixed circular orbits, called shells, around the nucleus; every other radius is forbidden.2. Each orbit carries a fixed, definite energy, and while confined to one orbit an electron neither absorbs nor radiates energy. \[\text{electron confined to an orbit} \Rightarrow E_n = \text{constant (stationary state)} \]3. An electron gains or loses energy only on jumping between orbits, equal to the orbit-energy difference. \[E_2 - E_1 = h\nu \]Answer: fixed, definite-energy orbits (stationary states); no radiation within an orbit; energy exchanged, \(\displaystyle \Delta E = h\nu\), only on jumping orbits.

  7. Exercise 42

    Show diagramatically the electron distributions in a sodium atom and a sodium ion and also give their atomic number.

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    NCERT’s answer
    Since the atomic number of sodium atom is $\displaystyle 11$, it has $\displaystyle 11$ electrons. A positively charged sodium ion (\(\displaystyle Na^{+}\)) is formed by the removal of one electron from a sodium atom. So, a sodium ion has $\displaystyle 11$-$\displaystyle 1$ = $\displaystyle 10$ electrons in it. Thus, electronic distribution of sodium ion will be $\displaystyle 2$, 8. The atomic number of an element is equal to the number of protons in its atom. Since, sodium atom and sodium ion contain the same number of protons, therefore, the atomic number of both is 11.
    Sodium atom (Na): \[Z_{\text{Na}} = 11 \] \[2n^2:\ K_{\max}=2,\ L_{\max}=8,\ M_{\max}=18 \] \[K=2,\quad L=8,\quad M = 11-2-8 = 1 \]Sodium ion (\(\displaystyle \text{Na}^+\)): loses the lone M-shell electron, reaching a stable octet. \[K=2,\quad L=8,\quad M=0,\quad \text{total electrons}=10 \]NCERT_Solution_Class9_Science_Exemplar_Ch4_Q42Atomic number counts protons, unaffected by ionisation. \[Z_{\text{Na}} = Z_{\text{Na}^+} = 11 \]Answer: Na: K $\displaystyle 2$, L $\displaystyle 8$, M $\displaystyle 1$; \(\displaystyle \text{Na}^+\): K $\displaystyle 2$, L $\displaystyle 8$; atomic number \(\displaystyle =11\) for both.

  8. Exercise 43

    In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherford’s model of an atom, ~ 1.00\displaystyle 1.00% of the α-particles were found to deflect at angles > 50\displaystyle 50º. If one mole of α-particles were bombarded on the gold foil, compute the number of α-particles that would deflect at angles less than 500\displaystyle 50^{0}.

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    % of α-particles deflected more than \(\displaystyle 50^{0}\) =$\displaystyle 1$% of α-particles. % of α-particles deflected less than \(\displaystyle 50^{0}\) =$\displaystyle 100$-$\displaystyle 1$ = $\displaystyle 99$% Number of α-particles bombarded = $\displaystyle 1$ mole = $\displaystyle 6.022$×\(\displaystyle 10^{23}\) particles Number of particles that deflected at an angle less than \(\displaystyle 50^{0}\) = × $\displaystyle 6.022$ ×$\displaystyle 10$ $\displaystyle 596.178$ = ×$\displaystyle 10$ = $\displaystyle 5.96$ ×$\displaystyle 10$ Sodium atom Sodium ion
    Deflected at \(\displaystyle >50^\circ\) and at \(\displaystyle <50^\circ\) together make up the whole mole. \[N = 1\ \text{mol} = 6.022\times10^{23}\ \text{particles} \] \[\%_{<50^\circ} = 100\% - 1.00\% = 99.00\% \] \[N_{<50^\circ} = 0.9900 \times 6.022\times10^{23}\ \text{particles} \] \[N_{<50^\circ} = 5.96\times10^{23}\ \text{particles} \]Answer: \(\displaystyle 5.96\times10^{23}\) \(\displaystyle \alpha\)-particles deflect at less than \(\displaystyle 50^\circ\).