Atoms MCQs for NEET — Physics Questions with Answers

When continuous light from a source is examined directly in a spectroscope, we observe the emission spectrum of the source. The sodium vapour spectrum consists of a few isolated bright lines. Each bright-line corresponds to a particular wavelength. It is emitted by the atoms in the gaseous state.

When continuous light from a source is made to pass through an absorbing substance and then examined in a spectroscope, we observe the absorption spectrum of the substance.

A band spectrum is emitted by chemical compounds in the vapour state. It is therefore a molecule spectrum.

A continuous emission spectrum consists of a wide range of unseparated wavelengths.

Radii of Bohr's stationary orbit is given by

$$\begin{gathered} I=\frac{n^2{h}^2}{4{\mathrm{\pi }}^2{\mathrm{mke}}^2\mathrm{Z}} \\ \therefore \mathrm{I} \mathrm{\alpha } \frac{{\mathrm{n}}^2}{\mathrm{Z}} \end{gathered}$$

Considering two situations of electrons,

$$\begin{gathered} \Rightarrow \frac{\left(r_f\right)}{\left(r_i\right)}=\frac{n_f^2}{n_i^2} \\ For ground state n_i=1 \\ \therefore \frac{21.2\times {10}^{-11}}{5.3\times {10}^{-11}}=n_f^2 \\ or n_f^2=4 \\ \therefore n_f=2 \end{gathered}$$

From Bohr's postulate, for any permitted (stationary orbit). Angular momentum of electron revolving in an orbit is constant

i.e. $$\begin{gathered} mvr=\frac{nh}{2\mathrm{\pi }} \\ \mathrm{or} \mathrm{v}=\frac{\mathrm{nh}}{2\mathrm{\pi mr}}.......\left(\mathrm{i}\right) \\ \mathrm{Also}, \frac{{\mathrm{mv}}^2}{\mathrm{r}}=\frac{{\mathrm{Ze}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}=\frac{{\mathrm{kZe}}^2}{{\mathrm{r}}^2}.....\left(\mathrm{ii}\right) \\ (\mathrm{where}, \mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}) \end{gathered}$$

Symbols have their usual meaning. From Eqs (i) and (ii)

$$\begin{gathered} r=\frac{n^2{h}^2}{4{\mathrm{\pi }}^2{\mathrm{mkZe}}^2} \\ \mathrm{For} \mathrm{hydrogen} \mathrm{atom}, \\ \mathrm{Z}=1 \\ \therefore \mathrm{r}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{mke}}^2} \\ \Rightarrow {\mathrm{r}}_{\mathrm{n} }\mathrm{\alpha } {\mathrm{n}}^2 \\ \therefore {\mathrm{a}}_2=4{\mathrm{a}}_0 \end{gathered}$$

$$\begin{gathered} {\lambda }_{min}=\frac{12400}{V}\stackrel{0}{A}=\frac{12400}{30000} \\ =0.413 \stackrel{0}{A} \end{gathered}$$

E_n = $-\frac{13.6}{n^2} eV ;$ 

$E_{n_2}- E_{n_1} = +13.6\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)eV$

This is maximum for n₁ = 1, n₂ = 3. (Choice B)

$Bohr radius r=\frac{{\epsilon }_0{n}^2{h}^2}{\pi Zm{e}^2}; \therefore r \alpha n^2$

According to Bohr's hypothesis, electron can revolve only in those orbits in which its angular momentum is an integral multiple of $\frac{h}{2\mathrm{\pi }}$, where h is Plank's constant.  In these orbits, angular momentum of electron can gave magnitude as $\frac{h}{2\mathrm{\pi }}, \frac{2h}{2\mathrm{\pi }}, \frac{3h}{2\mathrm{\pi }}....$etc, but never as $\frac{1.5 h}{2\mathrm{\pi }}, \frac{2.5 h}{2\mathrm{\pi }}, \frac{3.5 h}{2\mathrm{\pi }}....$etc.  This is called the quantisation of angular momentum.

When a hydrogen atom is raised from the ground state to excited state Potential energy increases and KE decreases, because

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ where \frac{m{v}^2}{r_n}=\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}^2} \\ m{v}^2=\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}} \\ \therefore KE=\frac{1}{2}\left(\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}}\right)=\frac{Z{e}^2}{8{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}} \\ PE=-\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}} [-ve of twice the KE] \end{gathered}$$

When the electron jumps from n = 1 to n = 2

since $r_n \alpha n^2$ radius of orbit increases

$\therefore KE decreases$

Since PE is -ve, when $r_n$ increases, $\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}}$ decreases - $\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}}$ increases

So PE increases, KE decreases,

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For Paschen series, $n_1=3$ for extreme lines of Paschen series $n_2=4$ and $n_2=\infty$

$$\begin{gathered} \therefore \frac{1}{{\lambda }_1}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=1.097\times {10}^7(0.0486) \\ {\lambda }_1=1.88\times {10}^{-6}m=1.88 \mu m \\ and \frac{1}{{\lambda }_2}=R\left(\frac{1}{3^2}-\frac{1}{{\infty }^2}\right)=1.097\times {10}^7\left(\frac{1}{9}\right) \\ {\lambda }_2=8.2\times {10}^{-7}m \\ =0.82\times {10}^{-6}m=0.82 \mu m \end{gathered}$$