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 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.

Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

If the IP of hydrogen in its ground state is 2.18 x$10^{-18}$ J/atom, then the electron affinity of $Li^{3+}$ ion is :

  1. $-2.18\times$$10^{-18}$J/atom
  2. $-6.54 \times$$10^{-18}$J/atom
  3. $-3.488 \times$$10^{-18}$J/atom
  4. $-1.962\times$$10^{-17}$ J/atom
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

E.A of $Li^{+3} = -$I.P of $Li^{+2}$
I.P of $Li^{+2}  = $I.P of $H \dfrac{Z^2}{n^2}$
Therefore, I.P of $Li^{+2}$=$2.18 \times 10^{-18} J/atom \times 9 $ as Z=3
Thus, E.A of $Li^{+3} =-1.962 \times 10^{-17}$ J/atom

Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

A satellite sent into space samples the density of matter within the solar system and gets a value $2.5$ hydrogen atoms per cubic centimeter. What is the mean free path of the hydrogen atoms? Take the diameter of a hydrogen atoms as $d=0.24\ nm$.

  1. $1.56\times 10^{12}\ m$
  2. $2.56\times 10^{12}\ m$
  3. $3.56\times 10^{12}\ m$
  4. $4.56\times 10^{12}\ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mean free path formula is lambda = 1 / (sqrt(2) * pi * n * d^2). Given n = 2.5 * 10^6 m^-3 and d = 0.24 * 10^-9 m, calculating this yields approximately 1.56 * 10^12 m.

Multiple choice physics constellations and galaxies light year evolution and end stages of stars in the world of stars

The $6563 A^0 H _\alpha$ line emitted by hydrogen in a star is found to be red-shifted by $15 A^0$. The speed with which the star is receding from the earth is

  1. $3.2 \times 10^5m s^{-1}$
  2. $6.87 \times 10^5m s^{-1}$
  3. $2 \times 10^5m s^{-1}$
  4. $12.74 \times 10^5m s^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Wavelength of $H _\alpha$ line, 
$\lambda=6563A^o$
$=6563\times10^{-10}$


Redshift observed in star $(\acute{\lambda-\lambda})=15A^o=15\times 10^{-10}$

Let the velocity of the star with which it is receding away from the earth be v.
Red shift relation, $\acute{\lambda}-\lambda$=$\dfrac{v}{c}\lambda$

$=\dfrac{c}{\lambda}\times (\acute{\lambda}-\lambda)$

$=\dfrac{3\times 10^8\times 15\times 10^{-10}}{6563\times 10^{-10}}$
$=6.87 \times 10^5m s^{-1}$

Multiple choice physics constellations and galaxies light year evolution and end stages of stars in the world of stars

The $6563 \mathring {A}$ line emitted by hydrogen atom in a star is found to be red shifted by $5\mathring{A}$. the speed with which the star is receding from the earth is:

  1. $17.3\times 10^3 $ m/s
  2. $4.29 \times 10^7 $m/s
  3. $3.39 \times 10^5 $m/s
  4. $ 2.29 \times 10^5$ m/s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac{\triangle v}{v}=\dfrac{V _{radiant}}{C}$
$v=\dfrac{C}{\lambda}$
$d.v=\dfrac{C}{\lambda^2}.dv=\dfrac{\dfrac{x}{\lambda^2}}{\dfrac{C}{\lambda}}=\dfrac{d\lambda}{\lambda}$
$\dfrac{d\lambda}{\lambda }=\dfrac{V _{radial }}{C}$
$\dfrac{5 \times 10^{-10}}{6563 \times 10^{-10}}=\dfrac{C _{radial}}{3\times 10^8}$
Calculate $V _radial=\dfrac{5}{6563}\times 3\times 10^8$
$=\dfrac{15}{6563}\times 10^8$
$\dfrac{15}{6.56}\times 10^5$
$2.29\times 10^5m/s$

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

The excitation energy of a hydrogen like ion to first excited state is 40.8 eV. The energy needed to remove the electron from the ion the ground state is

  1. 54.4 eV

  2. 62.6 eV

  3. 72.6 eV

  4. 58.6 eV.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a hydrogen atom like system the energy for $n^{th}  $ level is given by $E _n=\dfrac{-13.6Z^2}{n^2}eV$

So the ground state energy will be $E _1=\dfrac{-13.6Z^2}{1}eV$
and for the state $n=2$ the energy is $E _2=\dfrac{-13.6Z^2}{4}eV$
So the excitation energy foe this state will be $E _2 -E _1=10.2Z^2eV$
as it is given to be $40.8eV$  so $Z^2=4$ or $Z=2$
So the energy in ground state will be $E _1=-54.4eV$
so the inonization energy will be $-(-54.4eV)=54.4eV$

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

A free hydrogen atom in ground state is at rest. A neutron of kinetic energy K collides with the hydrogen atom. After collision hydrogen atom emits two photons in succession one of which has energy $2.55eV$. Assume that the hydrogen atom and neutron has same mas. 

  1. Minimum value of K is $25.5eV$
  2. Minimum value of K is $12.75eV$
  3. The other photon has energy $10.2eV$
  4. Th upper energy level is of excitation energy $12.5eV$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To excite hydrogen to the n=3 level (12.09 eV), the neutron must provide at least that much energy. The emission of 2.55 eV corresponds to n=4 to n=2 transition. The minimum energy required to reach n=4 is 12.75 eV.

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

If the ionization energy of hydrogen atom is $13.6 eV$ then the wavelength of the radiation required to excite the electron in $L{ i }^{ ++ }$ from first to third Bohr orbit is approximately

  1. $1140 A$
  2. $914 A$
  3. $11.4 A$
  4. $134 A$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ionisation energy is given by

$E = 13.6\ eV$ (given)
or $E = 13.6 \times 1.6 \times 10^{-19} J ....(1)$
also $E = hv$
$E = \dfrac{hc}{\lambda} ....(2)$
equation $(1)$ and $(2)$
$\dfrac{hc}{\lambda} = 13.6 \times 1.6 \times 10^{-19}$
$\lambda = \dfrac{h \times c}{13.6 \times 1.6 \times 10^{-19}}$
$\lambda = \dfrac{3 \times 10^{8} \times 6.63 \times 10^{-34}}{13.6 \times 1.6 \times 10^{-19}}$
$\lambda = 914 \times 10^{-10} m$
$\lambda = 914 A^o$

Multiple choice physics our solar system planets of the solar system solar system and sun introduction to solar system

Degenerate electron pressure will not be sufficient to prevent the core collapse of white dwarf if its mass becomes n times of our solar mass. Value of n is:

  1. 0.5

  2. 0.8

  3. 1

  4. 1.4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is the definition of Chandrasekhar's Limit . 
Degenerate electron pressure will not be sufficient to prevent core collapse of white dwarf if its mass becomes 1.4 times or more of our solar mass.

Multiple choice physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

The Schrodinger equation for a free electron of mass m and energy E written in terms of the wave function $\psi $ is $\frac{d^2\psi}{dx^2}+\frac{8 \pi ^2mE}{h^2}\psi =0$. The dimensions of the coefficient $\psi$ of in the second term must be

  1. $[M^1L^1]$
  2. $[L^2]$
  3. $[L^{-2}]$
  4. $[M^1L^{-1}T^1]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By dimensional analysis the dimensions of each term in an equation must be the same. In the first term the second derivative with respect to distance x indicates the dimensions of the coefficient $\psi$ of to be $[L^{-2}]$ and hence the answer.