Atoms MCQs for NEET — Physics Questions with Answers

Thomson model

$ {1 \over \lambda } = R2^2[{1\over n_2^2} -{1\over n_1^2}] \leftarrow for \, any \,atom $

$ { 1 \over \lambda_1 (L) } = R(1)^2[{1\over 1^2} - {1 \over 2^2}]=RZ^2[{1\over 4} - { 1 \over 16 } ] {3RZ^2 \over 4}$

$ { 1 \over \lambda_1 (L) } = {3R \over 4} \Rightarrow \lambda_1(L) = { 4 \over 3R} $

$ { 1 \over \lambda_2 (B) } = R2^2[{1\over 2^2} - {1 \over 4^2}]=RZ^2[{1\over 4} - { 1 \over 16 } ] = {3RZ^2 \over 16}$

$ \Rightarrow \lambda _2 (B) = { 16 \over 3RZ^2} $ $ { \lambda_1 (L) \over \lambda _2 (B) }$ $ 1 = { 2^2 \over 4} \Rightarrow z^2= 4 Z=2$

$ {1 \over \lambda} R [{ 1 \over 1 ^2 } - { 1 \over n^2 } ] $ $ {1 \over \lambda } = R [ 1- { 1 \over n^2 } ]$ $ { 1 \over \lambda } = R - { R/ n^2}$ $ { R \over n^2 }= R -{1/ \lambda } = { \lambda R -1 \over \lambda }$ $ { R \over n^2 } = { \lambda R -1 \over \lambda}$ $ n^2 = { \lambda R \over \lambda R -1 }$ $ n = \sqrt {\lambda R \over \lambda R -1}$

$$\begin{gathered} \overrightarrow{F}=-k\frac{e^2}{r^2}\hat{r} \left(Vector form\right) \\ =-k.\frac{e^2}{r^3}\overrightarrow{r} \end{gathered}$$    $\left(\because \hat{r}=\frac{{\displaystyle \overrightarrow{r}}}{{\displaystyle r}}\right)$  

$$\begin{gathered} \mathrm{In} \mathrm{ground} \mathrm{state} \mathrm{n}=1 \mathrm{and} \mathrm{for} \mathrm{first} \mathrm{exited} \mathrm{state} \mathrm{n}=2 \\ \mathrm{KE}=\frac{1}{4\mathrm{\pi } {\in }_0}\frac{{\mathrm{e}}^2}{2\mathrm{r}}\left(\mathrm{z}=1\right) =\frac{14.4\times {10}^{-10}}{2\mathrm{r}}\mathrm{eV} \left(\because \mathrm{r}=0.53{\mathrm{n}}^2 \stackrel{\circ }{\mathrm{A}}, \left(\mathrm{z}=1\right)\right) \\ {\left(\mathrm{KE}\right)}_1=\frac{14.4\times {10}^{-10}}{2\times 0.53\times {10}^{-10}}\mathrm{eV} =13.58 \mathrm{eV} \\ \mathrm{and} {\left(\mathrm{KE}\right)}_2=\frac{14.4\times {10}^{-10}}{2\times 0.53\times {10}^{-10}\times 4}\mathrm{eV}=3.39 \mathrm{eV} \\ \therefore \mathrm{KE} \mathrm{decreases} \mathrm{by} =10.2 \mathrm{eV} \\ \therefore \mathrm{PE} \mathrm{increases} \mathrm{by} = \mathrm{Excitation} \mathrm{energy} + \mathrm{Loss} \mathrm{in} \mathrm{kinetic} \mathrm{energy} = 10.2 + 10.2 = 20.4 \mathrm{eV} \\ \hspace{0.17em}\mathrm{Now} \mathrm{angular} \mathrm{momentum}, \mathrm{L}= \mathrm{mvr} = \frac{\mathrm{nh}}{2\mathrm{\pi }} \\ \Rightarrow {\mathrm{L}}_2-{\mathrm{L}}_1=\frac{\mathrm{h}}{2\mathrm{\pi }}=\frac{6.6\times {10}^{-34}}{6.28}=1.05\times {10}^{-34} \mathrm{J}‐\sec \end{gathered}$$

$\frac{k{e}^2}{r^2}=\frac{m{v}^2}{r} \Rightarrow \frac{1}{2} m{v}^2=\frac{k{e}^2}{2r}$

When hydrogen atom emits a photon in the Balmer series, electron transition occurs from n = 3, n = 4..... etc to n = 2. The electron transition from n = 2 to n = 1 also may take place. Therefore, it must emit a photon in Lyman series.

$E_n=-\frac{13.6}{n^2}eV; E_1=-13.6 eV$

To remove electron from n = 1 to n = $\infty$, 13.6 volt Potential (ionisation potential) is needed.

In line emission spectrum, every line spectrum consists of a few isolated bright lines, each bright line corresponds to a particular wavelength. It is emitted by atoms in the gaseous state.

e.g a sodium discharge lamp, a mercury vapour lamp, a neon discharge tube and a helium discharge tube all emit sharp lines of definite wavelength.

Ionisation energy corresponding to ionisation potential ($E_1=-13.6 eV$)

Photon energy incident ($∆E$) = 12.1 eV

So, the energy of electron in excited state ($E_2$) is given by

$$\begin{gathered} E_2-E_1=∆E \\ \Rightarrow E_2=∆E+E_1 \\ \Rightarrow E_2=-13.6+12.1 \\ \Rightarrow E_2=-1.5 eV \\ i.e. E_2=-\frac{13.6}{n^2}eV \\ -1.5=\frac{-13.6}{n^2} \\ \Rightarrow n^2=\frac{-13.6}{-1.5}\approx 9 \\ \therefore n=3 \\ i.e energy of electron in excited state corresponds to third orbit. \\ The possible spectral lines is given by \\ \frac{n(n-1)}{2} \\ \Rightarrow \frac{3(3-1)}{2} \\ \Rightarrow 3 \end{gathered}$$