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NCERT Exemplar · Class 12 Physics Atoms

29 questions

Short Answer Questions 19–23 (part 4 of 5)

  1. Exercise 19

    Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?
    NCERT’s answer
    The '\(\displaystyle m\)' that occurs in the Bohr formula \(\displaystyle E_n=-\frac{m e^4}{8 \varepsilon_0 n^2 h^2}\) is the reduced mass. For H-atom \(\displaystyle m \approx m_{\mathrm{e}}\). For positronium \(\displaystyle m \approx m_e / 2\). Hence for a positonium \(\displaystyle E_1 \approx-6.8 \mathrm{eV}\).
    The two particles have equal mass \(\displaystyle m\) and circle their common centre of mass, so the reduced mass \(\displaystyle \mu\) replaces \(\displaystyle m\), not the electron mass alone.\[\mu=\frac{m\cdot m}{m+m}=\frac{m}{2} \]\[E_n=-\frac{\mu e^{4}}{8\varepsilon_0^{2}h^{2}}\cdot\frac{1}{n^{2}},\qquad \mu_{\mathrm{H}}\approx m\;\Rightarrow\;E_1(\mathrm{H})=-13.6\ \mathrm{eV} \]\[\frac{E_1(\mathrm{Ps})}{E_1(\mathrm{H})}=\frac{\mu_{\mathrm{Ps}}}{\mu_{\mathrm{H}}}=\frac{m/2}{m}=\frac{1}{2} \]\[E_1(\mathrm{Ps})=\frac{1}{2}\times(-13.6\ \mathrm{eV})=-6.8\ \mathrm{eV} \]Answer: \(\displaystyle E_1=-6.8\ \mathrm{eV}\)

  2. Exercise 20

    Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstances, calculate the ground state energy of a He-atom.
    Each electron feels only the nucleus (\(\displaystyle Z=2\)); with no mutual repulsion both sit in \(\displaystyle n=1\), like the electron of \(\displaystyle \mathrm{He^{+}}\).\[E_1=-13.6\,\frac{Z^{2}}{n^{2}}\ \mathrm{eV}=-13.6\times\frac{4}{1}\ \mathrm{eV}=-54.4\ \mathrm{eV}\ \text{(per electron)} \]\[E_{\mathrm{He}}=2\times(-54.4\ \mathrm{eV})=-108.8\ \mathrm{eV} \]Caution: add no electron-electron term.Answer: \(\displaystyle E=-108.8\ \mathrm{eV}\)

  3. Exercise 21

    Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.
    The electron crosses any point of its orbit once per period \(\displaystyle T\), so a charge \(\displaystyle e\) passes per period.\[r_1=\frac{\varepsilon_0h^{2}}{\pi me^{2}}=0.529\times10^{-10}\ \mathrm{m},\qquad v_1=\frac{e^{2}}{2\varepsilon_0h}=2.19\times10^{6}\ \mathrm{m\,s^{-1}} \]\[T=\frac{2\pi r_1}{v_1}=\frac{2\pi\,(0.529\times10^{-10}\ \mathrm{m})}{2.19\times10^{6}\ \mathrm{m\,s^{-1}}}=1.52\times10^{-16}\ \mathrm{s} \]\[I=\frac{e}{T}=\frac{1.6\times10^{-19}\ \mathrm{C}}{1.52\times10^{-16}\ \mathrm{s}}\approx1.05\times10^{-3}\ \mathrm{A} \]Answer: \(\displaystyle I\approx1.05\ \mathrm{mA}\)

  4. Exercise 22

    Show that the first few frequencies of light that is emitted when electrons fall to the nth \displaystyle n^{\text {th }} level from levels higher than n\displaystyle n, are approximate harmonics (i.e. in the ratio 1:2:3…\displaystyle 1: 2: 3 \ldots) when n≫1\displaystyle n \gg 1.
    NCERT’s answer
    \(\displaystyle \nu_{\mathrm{mn}}=c R Z^2\left[\frac{1}{(n+p)^2}-\frac{1}{n^2}\right]\), where \(\displaystyle m=n+p,(p=1,2,3, \ldots)\) and \(\displaystyle R\) is Rydberg constant. For \(\displaystyle p \ll n\). \[\begin{aligned} & \nu_{m n}=c R Z^2\left[\frac{1}{n^2}\left(1+\frac{p}{n}\right)^{-2}-\frac{1}{n^2}\right] \\ & \nu_{m n}=c R Z^2\left[\frac{1}{n^2}-\frac{2 p}{n^3}-\frac{1}{n^2}\right] \\ & \nu_{m n}=c R Z^2 \frac{2 p}{n^3} \simeq\left(\frac{2 c R Z^2}{n^3}\right) p \end{aligned} \] Thus, \(\displaystyle \nu_{\mathrm{mn}}\) are approximately in the order $\displaystyle 1$, $\displaystyle 2$, 3..........
    Take the jump from level \(\displaystyle n+p\) down to \(\displaystyle n\), with \(\displaystyle p=1,2,3,\dots\) and \(\displaystyle p\ll n\). \(\displaystyle R\) is the Rydberg constant.\[\nu_p=Rc\left[\frac{1}{n^{2}}-\frac{1}{(n+p)^{2}}\right] \]\[\frac{1}{(n+p)^{2}}=\frac{1}{n^{2}}\left(1+\frac{p}{n}\right)^{-2}\approx\frac{1}{n^{2}}\left(1-\frac{2p}{n}\right) \]\[\nu_p\approx Rc\left[\frac{1}{n^{2}}-\frac{1}{n^{2}}+\frac{2p}{n^{3}}\right]=\frac{2Rc}{n^{3}}\,p \]\[\nu_1:\nu_2:\nu_3:\dots=1:2:3:\dots \]Caution: the expansion holds only while \(\displaystyle p\ll n\).Answer: \(\displaystyle \nu_p\approx 2Rc\,p/n^{3}\), so the frequencies are in the ratio \(\displaystyle 1:2:3\dots\)

  5. Exercise 23

    What is the minimum energy that must be given to a H atom in ground state so that it can emit an Hγ\displaystyle H_\gamma line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such Hγ\displaystyle H_\gamma photon?
    NCERT’s answer
    \(\displaystyle H_\gamma\) in Balmer series corresponds to transition \(\displaystyle n=5\) to \(\displaystyle n=2\). So the electron in ground state \(\displaystyle n=1\) must first be put in state \(\displaystyle n=5\). Energy required \(\displaystyle =E_1-E_5=13.6-0.54=13.06 \mathrm{eV}\). If angular momentum is conserved, angular momentum of photon = change in angular momentum of electron \[\begin{aligned} & =L_5-L_2=5 \hbar-2 \hbar=3 \hbar=3 \times 1.06 \times 10^{-34} \\ & =3.18 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} . \end{aligned} \]
    \(\displaystyle \mathrm{H}_\gamma\) is the \(\displaystyle n=5\to2\) line (\(\displaystyle \mathrm{H}_\alpha\) is \(\displaystyle 3\to2\), \(\displaystyle \mathrm{H}_\beta\) is \(\displaystyle 4\to2\)), so the atom must first be raised from \(\displaystyle n=1\) to \(\displaystyle n=5\).NCERT_Solution_Class12_Physics_Exemplar_Ch12_Q23\[E=E_5-E_1=13.6\left(1-\frac{1}{25}\right)\ \mathrm{eV}=13.6\times0.96\ \mathrm{eV}=13.06\ \mathrm{eV} \]With \(\displaystyle L_n=\dfrac{nh}{2\pi}\), the atom's loss of angular momentum goes to the photon:\[L_{\gamma}=L_5-L_2=\frac{(5-2)\,h}{2\pi}=\frac{3\times6.63\times10^{-34}\ \mathrm{J\,s}}{2\pi}=3.17\times10^{-34}\ \mathrm{J\,s} \]Answer: \(\displaystyle E=13.06\ \mathrm{eV}\); \(\displaystyle L_\gamma=\dfrac{3h}{2\pi}\approx3.17\times10^{-34}\ \mathrm{J\,s}\)