Physics · Chemistry

Atomic Structure and Spectra

113 Questions

Atomic structure and spectra questions cover the Bohr model, quantum numbers, and electron transition calculations. These topics are fundamental to the physics and chemistry syllabi of competitive exams. Solving these builds confidence in handling atomic physics problems.

Bohr atomic modelHydrogen spectrum seriesElectron binding energyQuantum numbersDe Broglie wavelength

Atomic Structure and Spectra Questions

Multiple choice bohr's model of atom structure of atom

The energy of second Bohr orbit of the hydrogen atom is $-328kJ$ ${mol}^{-1}$. Hence the energy of fourth Bohr orbit would be:

  1. $-41kJ$ ${mol}^{-1}$
  2. $-13121kJ$ ${mol}^{-1}$
  3. $-164kJ$ ${mol}^{-1}$
  4. $-82kJ$ ${mol}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
According to the Bohr model, the energy of the orbit is inversely proportional to the square of orbit number:

$E\quad \alpha \quad 1/n²$ -------- (1)

$\dfrac{E _1}{E _2} = \dfrac{n _2}{n _1}$

$\dfrac{-328}{E _2} = \dfrac{4^2}{2^2}$

$E _2 =\dfrac{-328kJ/mol}4$

$=-82kJ/mol.$

Hence, $-82kJ/mol$ is the answer.
Multiple choice bohr's model of atom structure of atom

The shortest $\lambda$ for the Lyman series of hydrogen atom is:


[Given, $R _H=109678 cm^{-1}]$

  1. $911.7A^o$
  2. $700 A^o$
  3. $600 A^o$
  4. $811 A^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
For Lyman series, $n _1=1$

For shortest $\lambda$ of Lyman series; energy difference in two levels showing transition should be maximum i.e $n _2= \infty$

$\dfrac{1}{\lambda}$$=R _H \left[\dfrac{1}{{1}^2}− \dfrac{1}{{(\infty})^2} \right]$

$\dfrac{1}{\lambda}$$=109678$

$\lambda$ $=911.7\times 10^{−8}$ cm

   $=911.7A^0$

Hence, option A is correct.
Multiple choice bohr's model of atom structure of atom

The first emission line in the atomic spectrum of hydrogen in the Balmer Series appears at:

  1. $\dfrac {9R _H}{400}cm^{-1}$
  2. $\dfrac {7R _H}{144}cm^{-1}$
  3. $\dfrac {3R _H}{4}cm^{-1}$
  4. $\dfrac {5R _H}{36}cm^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac {1}{\lambda}=R _HZ^2[\dfrac {1}{n _1^2}-\dfrac {1}{n _2^2}]$
For Balmer Series' first emission line,
$\dfrac {1}{\lambda}=R _H\times 1[\dfrac {1}{2^2}-\dfrac {1}{3^2}]=\dfrac {R _H5}{36}cm^{-1}$

Multiple choice bohr's model of atom structure of atom

Statement I : Wavelength of limiting line of lyman series is less than wavelength of limiting line of Balmer series.
Statement II: Rydberg constant value is same for all elements

  1. Statement I is true, Statement II is also true; Statement is the correct explanation of Statement I

  2. Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I

  3. Statement I is true, Statement II is false

  4. Statement I is false, Statement II is true
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

                              Lyman series                           Balmer series
Limiting line   :         $n=\infty :to:n=1$               $n=\infty :to:n=2$

                                                    $E _l > E _B$                           [limiting case]
                                                    $\lambda _l < \lambda _B$
Rydberg constant represents the limiting value of the highest wavenumber of any photon that can be emitted from an atom.
                                                    $\frac{1}{\lambda _l }=Z^2R        [\lambda _l =limiting :case]$

Multiple choice bohr's model of atom structure of atom

Which are correct for emission spectra of Balmer series in $H$-atom?

  1. $\displaystyle\lambda _{(in nm)}=364.56\left[\frac{n^2 _2}{n^2 _2-n^2 _1}\right]$; where $n _1=2$ and $n _2 > 2$
  2. $\dfrac{1}{\lambda}=R\left[\displaystyle\frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right];$ where $n _1=2$ and $n _2 > 2$; $R=3.29\times 10^{15}H _z$
  3. $\displaystyle\frac{1}{\lambda}=R _H \left[ \frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right ];$ where $n _1=2$ and $n _2 > 2$; $R _H=1.09737\times 10^5cm^{-1}$
  4. $\dfrac{1}{\lambda}=\frac{4c(in msec^{-1})}{364.56\times 10^{-9}}\left[\frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right];$ where $n _1=2$ and $n _2 > 2$;

    $c$ is speed of light.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

From Bohr model, we know that
in a transition,
$\displaystyle\frac{1}{\lambda}=R _H \left[ \frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right ]$ where, $R _H = 109700cm^-$
For Balmer series, electron gets deexcited from 3rd or upper level to second level so $n _1 =2, n _2 >2$.
Also we know that,
$E=hc/\lambda$ so
$\displaystyle v=\frac{4c(in msec^{-1})}{364.56\times 10^{-9}}\left[\frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right];$ where $n _1=2$ and $n _2 > 2$; $c$ is speed of light.

Multiple choice bohr's model of atom structure of atom

An $e^{-}$ of $He^{+}$ makes a transition and emits $6^{th}$ line of Balmer series. Similar wavelength of radiation is absorbed by hydrogen like specie to give $9^{th}$ line of paschen series in its spectrum. The value of Z of the hydrogen like specie is :

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Balmer $6^{th}$ line (8---2)
Paschen $9^{th}$ line (12---3)
For $\displaystyle He^{+}\frac{1}{\lambda }=R.2^{2}\left ( \frac{1}{2^{2}}-\frac{1}{8^{2}} \right )$

For $He$ atomic number is 2.
$\displaystyle \frac{1}{\lambda }=R\left ( \frac{1}{(2/2)^{2}}-\frac{1}{(8/2)^{2}} \right )$ ....(1)
Fo\displaystyle r single electron species having atomic number 't'
$\displaystyle \frac{1}{\lambda }=Rt^{2}\left ( \frac{1}{3^{2}}-\frac{1}{12^{2}}\right )$
Here t corresponds to atomic number Z of the element.
$\displaystyle \frac{1}{\lambda }=R\left ( \frac{1}{(3/t)^{2}}-\frac{1}{(12/t)^{2}} \right )$ ...(2)
Comparing (1) & (2)
$\Rightarrow \displaystyle \left ( \frac{2}{2} \right )^{2}=\left ( \frac{3}{t} \right )^{2}$
$\Rightarrow Z=3$

Multiple choice bohr's model of atom structure of atom

In a mixture of $H-He^{+}$ gas, H atom and $He^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer its excitation energy to $He^{+}$ ions by collision.
If each hydrogen atom in the ground state is excited by absorbing photons of energy 8.4 eV, 12.09 eV, of energy, then assuming the Bohr model of an atom is applicable the number of spectral lines emitted is equal to:

  1. 5

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Energy of electron in $n^{th}$ shell:


$E _n=  \dfrac{–13.12\ Z^2}{n^2} eV$


$n= 1: -13.6 eV$
$n= 2: -3.4 eV$
$n= 3: -1.51 eV$
$n=4: -0.85 eV$

$E(n=3)- E(n=1)= 12.09 eV$

So if $H$ atoms are excited by $8.4eV$ and $12.09 eV$ then the electrons will reach to $n=3$ shell. Then the no. of emitted spectral lines will be equal to $3$.


Hence, the correct option is $(C)$.

Multiple choice bohr's model of atom structure of atom

The emission spectrum of hydrogen is found to satisfy the expression for the energy change $\triangle E$ (in joules) such that $\triangle E = 2.18\times 18^{-18}(\frac{1}{n _1^2}-\frac{1}{n _2^2})J$ where $n _1$= 1, 2, 3, .......and $n _2$ = 2, 3, 4. The spectral lines corresponds to Paschen series if :

  1. $n _1 = 1$ and $ n _2 = 2, 3, 4$
  2. $n _1 = 3$ and $ n _2 = 4, 5, 6$
  3. $n _1 = 1$ and $ n _2 = 3, 4, 5$
  4. $n _1 = 2$ and $ n _2 = 3, 4, 5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the emission spectra of hydrogen atom, Paschen series is the one where the transition from higher energy states to third energy state takes place.

i.e. $n _f=n _1=3$ and $n _i=n _2>3$
option B

Multiple choice bohr's model of atom structure of atom

What would be the wavelength and name of series respectively for the emission transition for H-atom if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm?

  1. 434 nm, Balmer

  2. 434 pm, Paschen

  3. 545 pm, Pfund

  4. 600 nm, Lyman
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The radii of the $n^{th}$ stationary state for a hydrogen-like specie is expressed as :

$r _n = \cfrac{n^2 a _0}{Z}$
where $Z=$atomic number and $a _0=52.9\ pm$ radius of Bohr orbit.

For hydrogen atom, Z=1
Given that transition is from orbit radius = 1.3225 nm to 211.6 pm 
Orbit with radius = 1.3225 nm=1322.5 pm

$r _n=52.9 \times n^2=1322.5$

$n^2=25$ or $n=n _i=5$ 

Similarly for Orbit with radius = 211.6 pm

$r _n=52.9 \times n^2=211.6$

$n^2=4$ or $n=n _f=2$

thus transition is from $n _i=5$ to $n _f=2$

Transition energy from $n _i\ to\ n _f$ is given as:
$\frac{1}{\lambda}=R _H[\cfrac{1}{n _f^2}-\cfrac{1}{n _i^2}]$

where $R _H=109677cm^{-1}$ and $n _i=5,n _f=2$

$\cfrac{1}{\lambda}=109677[\cfrac{1}{2^2}-\cfrac{1}{5^2}]\ cm^{-1}$

$\cfrac{1}{\lambda}=109677[\cfrac{1}{4}-\cfrac{1}{25}]\ cm^{-1}$

$\cfrac{1}{\lambda}=109677 \times 0.21\ cm^{-1}$

$\cfrac{1}{\lambda}=23032.17\ cm^{-1}$

$\lambda=4.342\times 10^{-5} cm=434.2\ nm$
Since transition is from higher energy state to n=2, it belongs to Balmer series and wavelength of the transition is 434 nm

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

In Bohr's atom the number of de Broglie's waves associated with an electron moving in $n^{th}$ permitted orbit is:-

  1. $n$
  2. $2n$
  3. $\dfrac{n}{2}$
  4. $n^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
According to Bohr's postulate and de-Broglie the relation for $n^{th}$ orbit is like this $2\pi r _n=n\lambda$
Where $r _n$ is radius of $n^{th}$ orbit and $\lambda$ is deBroglie wavelength of electron in that orbit.
So number of de-Broglie wavelength is $n$.
Correct option is A.
Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

One of Bohr's assumptions about stable electron orbits in a hydrogen atom treated electrons as particles with circular orbits. Louis de Broglie made a different assumption about the electron and its stable orbits which turned out to be mathematically equivalent to Bohr' assumption.
What assumption did the Broglie make?

  1. The electron orbits the nucleus with elliptical-shaped orbits, like the planets around the sun

  2. The electron forms standing wave patterns that must fit a circular shape around the nucleus

  3. The electron follows a parabolic trajectory around the nucleus

  4. The electron behaves like a cloud, blocking out all radiation except radiation associated with allowable energy transitions

  5. The electron follows a hyperbolic trajectory around the nucleus
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A major problem with Bohr's model was that it treated electrons as particles that existed in precisely-defined orbits. de Broglie made a different assumption about the electron and its stable orbits which turned out to be mathematically equivalent to Boh'r assumption. He assumed that the electron forms standing wave patterns that must fit a circular shape round the nucleus.

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two gas jars kept separately at temperature $27^{o}C$ and $127^{o}C$ respectively is

  1. $\dfrac{2}{\sqrt{3}}$
  2. $2:3$
  3. $\dfrac{\sqrt{3}}{4}$
  4. $\sqrt{\dfrac{8}{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

de-Broglie wavelength $\lambda=\dfrac{h}{mv}$
Where the speed ($r.m.s$) of a gas particle at the given temperature ($T$) is given as $\dfrac{1}{2} mv^{2}=\dfrac{3}{2}kT$
$\Rightarrow \quad v=\sqrt{\dfrac{3KT}{m}}$ where $k=$ Boltzmanns constant and $m=$ mass of the gas particle and $T=$ temperature of the gas in $k$
$\Rightarrow \quad mv=\sqrt{3mKT}$
$\Rightarrow \quad \lambda=\dfrac{h}{mv}=\dfrac{h}{\sqrt{3mkT}}$
$\therefore \quad \dfrac{\lambda _{H}}{\lambda _{He}}=\sqrt{\dfrac{m _{He}\ T _{He}}{m _{H}\ T _{H}}}$
$=\sqrt{\dfrac{(4\ amu)\ (273 + 127)^{\circ }k}{(2\ amu)\ (273 + 127)^{\circ }k }}=\sqrt{\dfrac{8}{3}}$
Hence ($D$) is correct.

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

Imagine an atom made of a proton and a hypothetical partical of double the mass as that of an electron but the same charge. Apply Bohr theory to consider transitions of the hypothetical particle to the ground state. Then, the longest wavelength (in terms of Rydberge constant for hydrogen atom) is

  1. $\dfrac {1}{2R}$
  2. $\dfrac {5}{3R}$
  3. $\dfrac {1}{3R}$
  4. $\dfrac {2}{3R}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From the formula,

$\dfrac{1}{\lambda }={{R} _{new}}\left( \dfrac{1}{n _{1}^{2}}-\dfrac{1}{n _{2}^{2}} \right)$

Where,

Rydberg Constant,$R=\dfrac{m{{e}^{4}}}{8\varepsilon _{o}^{2}{{h}^{3}}c}$

New Constant, (for $m=2$ ),$\,{{R} _{new}}=\dfrac{\left( 2m \right){{e}^{4}}}{8\varepsilon _{o}^{2}{{h}^{3}}c}=2R$

Longest wavelength,

$ \dfrac{1}{{{\lambda } _{long}}}=2R\left( \dfrac{1}{n _{1}^{2}}-\dfrac{1}{n _{2}^{2}} \right)=2R\left( \dfrac{1}{1}-\dfrac{1}{{{2}^{2}}} \right)=\dfrac{6R}{4} $

$ \Rightarrow {{\lambda } _{long}}=\dfrac{2}{3R} $

Hence, longest wavelength is $\dfrac{2}{3R}$

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

de - Broglie wavelength of an electron in the nth bohr orbit is $\lambda _n$ and the angular momentum is $J _n$ then:

  1. $J _n\alpha \lambda _n$
  2. $\lambda _n \infty \dfrac{1}{J _n}$
  3. $\lambda _n \infty J _n^2$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$J=\dfrac{nh}{2\pi}=mvr$
$J=mvr,\lambda =\dfrac{h}{mv}$
$v\alpha \dfrac{1}{n} ,r\alpha n^2$
$\Rightarrow \lambda \alpha n$
$J\alpha n$
$\Rightarrow J\alpha \lambda$