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 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
B Correct answer
Multiple choice electromagnetic spectrum electromagnetic waves physics

A gas of identical hydrogen like atoms has some atoms in ground state and some atoms in a particular excited state and there are no atoms in any other energy level. The atoms of the gas make transition to a higher state by absorbing monochromatic light of wavelength $304 \AA $. subsequently, the atoms emit radiation of only six different photon energies. Some of emitted photons have wavelength $304 \AA $, some have wavelength more and some have less than $304 \AA $ ( Take $hc = 12420 eV - \AA $)
Find the principal quantum number of the initially excited state.

  1. 1

  2. 2

  3. 3

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

The energy of the absorbed photon is E = hc/lambda = 12420/304 approx 40.8 eV. For a hydrogen-like atom, E = 13.6 * Z^2 * (1/n1^2 - 1/n2^2). Given the emission of 6 lines, the final state must be n=4 (since 4*3/2 = 6). The transition is from n_initial to n_final = 4. Solving for Z and n_initial requires specific energy level data.

Multiple choice electromagnetic spectrum electromagnetic waves physics

A gas of identical hydrogen like atoms has some atoms in ground state and some atoms in a particular excited state and there are no atoms in any other energy level. The atoms of the gas make transition to a higher state by absorbing monochromatic light of wavelength $304 \mathring A $. subsequently, the atoms emit radiation of only six different photon energies. Some of emitted photons have wavelength $304 \mathring A $, some have wavelength more and some have less than $304 \mathring A $ ( Take $hc = 12420 eV - \mathring A $). Find the principal quantum number of the initially excited state.

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice electromagnetic spectrum electromagnetic waves physics

A gas of identical hydrogen like atoms has some atoms in ground state and some atoms in a particular excited state and there are no atoms in any other energy level. The atoms of the gas make transition to a higher state by absorbing monochromatic light of wavelength $304 \mathring A $. subsequently, the atoms emit radiation of only six different photon energies. Some of emitted photons have wavelength $304 \mathring A $, some have wavelength more and some have less than $304 \mathring A $ ( Take $hc = 12420 eV - mathring A $). Find the principal quantum number of the initially excited state.

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

In a hydrogen atom, an electron moves in an orbit of radius $5.0\times 10^{-11}\ m$ with a speed of $2.2\times 10^{6}\ ms^{-1}$. The equivalent current is :

  1. $11.2\times 10^{-3}\ A$
  2. $1.9\times 10^{-3}\ A$
  3. $1.12\times 10^{-3}\ A$
  4. $11.2\times 10^{4}\ A$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Current $=$ Charge passing per unit time $q$
$I=\dfrac { q }{ t } $
$t=\dfrac { 2\pi r }{ v } $
$I=\dfrac { q }{ 2\pi r } .v=\dfrac { 1.6\times { 10 }^{ 19 }\times 2.2\times { 10 }^{ 6 } }{ 5\times { 10 }^{ -11 }\times 2n } $
$I=\dfrac { 1.6\times 2.2 }{ 5\times 2\pi  } \times { 10 }^{ -2 }=1.12\times { 10 }^{ -3 }A$
$\boxed { I=1.12mA } $
Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

A spherical drop of radius $10$ has absorbed $40$ electrons. The energy required to given an additional electron to it is.

  1. $9.2 \times 10^{-21} J$
  2. $5.7 \times 10^{-21} J$
  3. $9.21 \times 10^{-23} J$
  4. $4Ke^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know that,
Energy $=$ Potential $\times $ Charge
$E=\left( \dfrac { k\quad q }{ r }  \right) \times e$
$E=\dfrac { k\left( 40e \right) \left( e \right)  }{ 10 } $
$E=4k{ \left( e \right)  }^{ 2 }$
$\boxed { Energy=4{ ke }^{ 2 } } $
Multiple choice energy in wave motion oscillation and waves waves physics

The wave number of energy emitted when electron jumps from fourth orbit to seconds orbit in hydrohen in $20,497\ cm^{-1}$. The wave number of energy for the same transition in $He^{+}$ is

  1. $5,099\ cm^{-1}$
  2. $20,497\ cm^{-1}$
  3. $40,994\ cm^{-1}$
  4. $81,988\ cm^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,

Wave number of energy for hydrogen $=20497\,c{{m}^{-1}}$

Now, for hydrogen

$\dfrac{1}{{{\lambda } _{H}}}=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)=20497$

Now, for helium

  $ \dfrac{1}{\lambda }=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)\times {{2}^{2}} $

 $ \dfrac{1}{{{\lambda } _{He}}}=20497\times 4 $

 $ \dfrac{1}{{{\lambda } _{He}}}=81988\,c{{m}^{-1}} $

Hence, the wave number of energy for helium is $81988\ cm^{-1}$

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Hydrogen atom will be in its ground state,if its electron is in

  1. any energy level

  2. the lowest energy state

  3. the highest energy state

  4. the intermediate state
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Hydrogen atom has one electron. Depending upon the electron residing in which energy state hydrogen energy varies. So, for hydrogen atom to be in ground state electron should be in lowest energy state.

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The ground state energy of the electron in hydrogen atom is equal to :

  1. the ground state energy of the electron in $ { He }^{ + } $
  2. the first excited state energy of the electron in $ { He }^{ + } $
  3. the first excited state energy of the electron in $ { Li }^{ +2 } $
  4. the ground state energy of the electron in $ { Be }^{ +3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The energy levels of hydrogen-like ions are given by E = -13.6 * Z^2 / n^2. For H (Z=1, n=1), E = -13.6 eV. For He+ (Z=2), the first excited state is n=2, so E = -13.6 * (2^2) / (2^2) = -13.6 eV.

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The total energy of an electron in the ground state of hydrogen atom is $-13.6\space eV$. The potential energy of an electron in the ground state of $Li^{2+}$ ion will be

  1. $122.4\space eV$
  2. $-122.4\space eV$
  3. $244.8\space eV$
  4. $-244.8\space eV$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Therefore, for ${ Li }^{ 2+ }$ ion, Total energy is $- 13.6\times 9 = 122.4\ eV$
But, $-K.E= T.E= \dfrac { P.E }{ 2 } $
Therefore, $P.E$ is $-244.8\ eV$