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 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 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.

Multiple choice physics wave optics huygens wave theory and wavefront wave propagation (huygens' construction) theories on light wave behaviour

The wavelength of characteristic $X \text { -ray } K _ { \alpha }$ a line emitted by hydrogen like atom is $0.32 \mathrm { A } ^{ \circ }$. The wavelength of $K _ { \beta }$ line emitted by the same element is 

  1. $0.21 { \mathrm { A } }$
  2. $0.27 \mathrm { A }$
  3. $0.33 \mathrm { A }$
  4. $0.40 \hat { \mathrm { A } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Moseley's law and hydrogen-like atomic energy levels, the frequency of characteristic X-rays is proportional to (Z - b)^2 and inversely proportional to squared wavelength differences. Using the transition formula for K-alpha and K-beta lines, the ratio of their wavelengths can be calculated, resulting in approximately 0.27 angstroms for K-beta.

Multiple choice evs - i substances, objects and energy renewable resources alternative fuels and energy sources alternative sources of energy

The total energy of an electron is 3.555 MeV then its kinetic energy is

  1. $3.545 \mathrm { MeV }$
  2. $3.045 \mathrm { MeV }$
  3. $3.5 \mathrm { MeV }$

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Every substance has energy due to its mass$:$

loss in energy
$\Delta E = \Delta m{c^2}$
$\therefore KE = \left( {m - {m _0}} \right){c^2}$
$ = 3.555 - {m _0}{c^2}$
$ = 3.555 - 0.51$
$ = 3.045MeV$
Hence,
option $(B)$ is correct answer.

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

The ionization potential for second He electron is 

  1. 13.6 V

  2. 27.2 V

  3. 54.4 V

  4. 3.4 V
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ionization energy of a hydrogen-like ion is given by E = 13.6 * Z^2 / n^2 eV. For the second electron of Helium (He+), Z = 2 and n = 1, so the energy is 13.6 * (2^2) / 1^2 = 13.6 * 4 = 54.4 eV.

Multiple choice physics spectra the electromagnetic spectrum electromagnetic spectrum electromagnetic waves

Neglecting reduced mass effects, what optical transition in the ${ He }^{ + }$ spectrum would have the same wavelength as the first Lyman transition of hydrogen ($n=2$ to $n=1$)

  1. $n = 2$ to $n = 1$
  2. $n = 3$ to $n = 1$
  3. $n = 4$ to $n = 2$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Rydberg formula for hydrogen-like ions is 1/lambda = R * Z^2 * (1/n1^2 - 1/n2^2). For hydrogen (Z=1), the Lyman transition n=2 to n=1 is 1/lambda = R(1 - 1/4) = 3R/4. For He+ (Z=2), the transition n=4 to n=2 gives 1/lambda = R * 2^2 * (1/4 - 1/16) = R * 4 * (3/16) = 3R/4. Thus, they have the same wavelength.

Multiple choice physics spectra the electromagnetic spectrum electromagnetic spectrum electromagnetic waves

The electron in hydrogen atom in a sample is in $n^{th}$ excited state, then the number of different spectrum lines obtained in its emission spectrum will be:

  1. $1 + 2 + 3 + ..... + (n - 1)$
  2. $1 + 2 + 3 + ..... + (n)$
  3. $1 + 2 + 3 + ..... + (n + 1)$
  4. $1$ x $2$ x $3$ x ..... x $(n - 1)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When an electron transitions from the n-th excited state (which corresponds to principal quantum number n + 1, or generally considering n as the level), the total number of spectral lines emitted during transitions to lower states is given by the sum from 1 to n.