NEET Practice Questions (MCQs) with Answers & Solutions

For Lyman series $\frac{1}{{\lambda }_{max}}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4}R$ and

 $$\begin{gathered} \frac{1}{{\lambda }_{min}}=R\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]=\frac{R}{1} \\ \Rightarrow \frac{{\lambda }_{max}}{{\lambda }_{min}}=\frac{4}{3} \end{gathered}$$

Number of spectral lines observed in hydrogen spectrum is given by

                         = $\frac{n(n-1)}{2} = \frac{4(4-1)}{2} = 6$

Where, n = principal quantum number = number of orbits.

Second line of Lyman series corresponds to the transition n = 3 $\to$n = 1

Second line of Balmer series corresponds to the transition n = 4 $\to$n = 2

Second line of Paschen series corresponds to the transition  n = 5 $\to$n = 3

An absorption line of Balmer series arises when electron jumps from n = 2 to any other higher state.

For, Brackett series

$n_2=5, 6, 7.....\to n_1=4$

For Pfund series

$n_2=6, 7, 8.....\to n_1=5$

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)

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