Physics · Chemistry

Atomic Structure and Spectra

149 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

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

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

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

Consider an electron in the ${ n }^{ th }$ orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of the de Brogile wavelength ${ n }^{ th }$ of the electron as :

  1. $(0.529)n\lambda $
  2. $\sqrt { n } \lambda $
  3. $(13.6)\lambda $
  4. $n\lambda $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By De-broglie hypotheseis -

$\begin{array}{l} mvr=\dfrac { { nh } }{ { 2\pi  } }  \ 2\pi r=n\left( { \dfrac { h }{ { mv } }  } \right) =n\lambda  \end{array}$

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

If the radius of first Bohrs orbit is $x$, then de-Broglie wavelength of electron in 3rd orbit is nearly

  1. $2\pi$ $x$
  2. $6\pi$$x$
  3. $9x$
  4. $x/3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Radius of 3rd orbit radius $=9x=n^{2}x$ (where $n=3$ )
Let de broglie wavelength be $\lambda $.
For the interference of the waves to be constructive,
$n\lambda =2\pi r$ ($r$ is radius of orbit)
$\Rightarrow \lambda =\dfrac{2\pi \times 9x}{3} $ (where, $\ n=3 $, the quantum state)
$\Rightarrow \lambda =6\pi x$

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

The circumference of the second orbit of an atom or ion having single electron ,is $4 \times10^{-9}$ m.The de-Brogile wavelength of electron revolving in this orbit should be

  1. $2\times 10^{-9}m$
  2. $4\times 10^{-9}m$
  3. $8\times 10^{-9}m$
  4. $1\times 10^{-9}m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circumference of the orbit $=4\times 10^{9}m$
The orbit number $=2$

$n\lambda =2\pi r$

$\Rightarrow \lambda =\dfrac{4\times 10^{9}}{2}m$

$\Rightarrow \lambda =2\times 10^{-9}m$

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

If the electron in hydrogen orbit jumps from third orbit to second orbit, the wavelength of the emitted radiation is given by

  1. $\lambda = \dfrac {R}{6}$
  2. $\lambda = \dfrac {5}{R}$
  3. $\lambda = \dfrac {36}{5R}$
  4. $\lambda = \dfrac {5R}{36}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that
$\dfrac {1}{\lambda} = R \left (\dfrac {1}{n _{1}^{2}} - \dfrac {1}{n _{2}^{2}} \right )$
$\dfrac {1}{\lambda} = R\left (\dfrac {1}{2^{2}} - \dfrac {1}{3^{2}} \right ) \Rightarrow R \left (\dfrac {1}{4} - \dfrac {1}{9}\right )$
$\dfrac {1}{\lambda} = \left (\dfrac {9 - 4}{36}\right ) R = \dfrac {5R}{36} \Rightarrow \lambda = \dfrac {36}{5R}$

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

According to de-Broglie explanation of Bohr's second postulate of quantization, the standing particle wave on a circular orbit for $n = 4$ is given by

  1. $2 \pi {r} _{n} = {4}/{\lambda}$
  2. $\dfrac{2 \pi}{\lambda} = 4{r} _{n}$
  3. $2 \pi {r} _{n} = 4 \lambda$
  4. $\dfrac{\lambda}{2 \pi} = 4 {r} _{n}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to debroglie explanation of Bohr's second postulate, assumption is made that integral number of wavelengths must fir in the circumference of circular orbit. The integral multiple comes out to be the same as quantization number.

$2 \pi r _n = n \lambda$
For $n=4$, 
       $2 \pi r _n = 4 \lambda$

Multiple choice laws of heat transfer heat and thermodynamics physics

The hydrogen atom in its ground state is excited by means of monochromatic radiation of energy $12.75ev$. How many different lines are possible in the resulting spectrum? You may assume the ionization energy for hydrogen atom as $13.6\ eV$

  1. $3$
  2. $4$
  3. $6$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The excitation energy is 12.75 eV. The energy levels of hydrogen are given by En = -13.6 / n^2 eV. The ground state (n=1) energy is -13.6 eV. The excited state energy is -13.6 + 12.75 = -0.85 eV. Solving for n gives -0.85 = -13.6 / n^2, so n^2 = 16, meaning n = 4. The number of spectral lines emitted when transitioning from n=4 to ground state is n(n-1)/2 = 4(3)/2 = 6.