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 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 dual nature of matter and radiation davisson and germer experiment and its conclusion matter waves wave nature of matter

In Davisson-Germer experiment electrons were made to strike a sheet made of ______ metal.

  1. Nickel

  2. Cobalt

  3. Iron

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

Davisson and Germer experiment was conducted by two American scientists Clinton Davisson and Lester Germer, in 1927, to verify the de Broglie hypothesis that a material particle posses wave nature. 

The electron beam was made to pass through a hole and strike the nickel crystal normally, the electrons scattered in all directions acting like waves. The detector indicated the peak intensity of scattered electrons at certain angle. This maximum intensity was due to constructive interference of two waves. Thus wave nature of electrons was experimentally proved.

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 $-244.8\ eV$

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 Hydrogen atom is $–13.6eV$. The potential energy of the electron in this state is:

  1. $0eV$
  2. $-27.2eV$
  3. $1eV$
  4. $2eV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ground state energy of hydrogen atom $=13.6eV$

Potential energy $=2$ energy of electron
                             $=2(-13.6\ eV)$
                             $=-27.2\ eV$

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

In which of the following systems will the radius of the first orbit (n = 1) be minimum ?

  1. hydrogen atom

  2. deuterium atom

  3. singly ionized helium

  4. doubly ionized lithium.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Radius of the first orbit of an atom  $R _1 = \dfrac{0.529}{Z}$  $A^o$
$\implies \ R _1 \propto\dfrac{1}{Z}$
where $Z$ is the atomic number of Hydrogen-like atom.
Since $Z$ is maximum for doubly ionized lithium, thus radius of first orbit is minimum in doubly ionized lithium.

Multiple choice chemistry periodic table electronic configuration and valency electron configuration periodic trends in physical properties

The ionisation potential of hydrogen atom is $13.6\ eV$. The energy of required to remove an electrons in the $n=2$ state of hydrogen atom is:

  1. $27.2\ eV$
  2. $13.6\ eV$
  3. $6.8\ eV$
  4. $3.4\ eV$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$E _n=-13.6\ ev\left (\dfrac {Z^2}{n^2}\right)=\dfrac {-E _1}{n^2}[I.E _1 =-E _1 =13.6\ eV \Rightarrow E _1=-13.6\ eV]$

$\therefore E _2 =\dfrac {-13.6\ ev}{4}=-3.4ev$

$\therefore I.E _2=-E _2 =3.4\ ev$.

Option D is correct.
Multiple choice physics units and measurement: error analysis accuracy of measurement accuracy and precision accuracy, precision and uncertainty in measurement

The accuracy in the measurement of the diametre of a hydrogen atom as ${ 1.06\times 10 }^{ -10 }$ m is

  1. $0.01$
  2. ${ 106\times 10 }^{ -10 }$
  3. $\dfrac { 1 }{ 106 } $
  4. ${ 0.01\times 10 }^{ -10 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Accuracy refers to how close a measurement is to the true value. The relative error is the absolute error divided by the measured value. Here, the precision is 0.01, so the accuracy is 0.01 / 1.06.

Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

The atoms of hydrogen combine to form a molecule of hydrogen gas, the energy of the $H _2$ molecule is:

  1. Greater than that of seperate atoms

  2. Equal to that of seperate atoms

  3. Lower than that of seperate atoms

  4. Some times lower and some times higher
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When atoms combine to form a stable molecule, energy is released, meaning the potential energy of the molecule is lower than that of the separate atoms.