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 physics wave motion reflection of waves

The wavelength of the first line of Lyman series is $\lambda$. The wavelength of the first line in Paschen series is ________.

  1. $108/7$
  2. $27/5$
  3. $7/108$
  4. $5/27$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Formula for Lyman series is:
Where, n = 2,3,4,5,.....
Where, R = Rydbergg constant,
$\dfrac{1}{\lambda}=R\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)$ and
 Paschen series in $\lambda _1$,
$\Rightarrow \dfrac{1}{\lambda _1}=R\left(\dfrac{1}{3^2}-\dfrac{1}{4^2}\right)$

$\Rightarrow \dfrac{\lambda _1}{\lambda}=\dfrac{\dfrac{3}{4}}{\dfrac{7}{16\times 9}}$

$\Rightarrow \lambda _1=\dfrac{3}{4}\times \dfrac{16\times 9}{7}\lambda$

$\Rightarrow \lambda _1=\dfrac{108}{7}\lambda$.

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

The $Z _{effective}$ for $He$ is?

  1. 2

  2. 1.7

  3. 1.85

  4. 1.65
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The effective nuclear charge experienced by a 1s electron in helium is +1.70.

The effective nuclear charge $Z _{eff}$ is the net positive charge experienced by an electron in a multi-electron atom.

A given electron does not experience a full nuclear charge because the other electrons are sometimes between it and the nucleus and shield it from the nucleus.

The formula for effective nuclear charge is-

$Z _{eff}=Z-S$

where, 
Z is the number of protons in the nucleus, and S is the shielding constant, the average number of electrons between the nucleus and the electron in question.
The American physicist John Slater derived a number of rules to determine the shielding constant.

He found that for electrons in a 1s orbital, the second electron shields the first by 0.30 units.

$Z _{eff}=Z-S=2- 0.30-1.70$

Hence, the correct option is B.


Multiple choice bohr's model of atom structure of atom

What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series? (h=Plank constant; C=Velocity of light; R=Rydberg constant)

  1. $\dfrac{5hcR}{36}$
  2. $\dfrac{4hcR}{3}$
  3. $\dfrac{3hcR}{4}$
  4. $\dfrac{7hcR}{144}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Use the formula,
$ 1/\lambda = R \times \left ( 1/n _1^2 - 1/n _2^2 \right) $.
Therefore,
$ \Delta E = hc/\lambda = hcR \times \left ( 1/n _1^2 - 1/n _2^2 \right) $.
for Lyman series lowest energy transition is from n$=1 $ to n$=2$.

Multiple choice bohr's model of atom structure of atom

The ratio of the wave numbers of the radiation corresponding to the third line of Balmer series and the second line of the Paschen series of hydrogen spectrum is:

  1. 21/16 x 9/4

  2. 25/16 x 9/4

  3. 21/25 x 9/4

  4. 16/25 x 9/4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Use the formula,
Wavenumber = $ R \times \left ( 1/n _1^2 - 1/n _2^2 \right ) $.
For third line Balmer series $ n _1=2$ and $n _2=5 $.
For second line Paschen series $ n _1=3$ and $n _2=5 $
Therefore the ratio is $ \left (1/4 - 1/25 \right) / \left (1/9 - 1/25 \right) $ which equals 
$ 21/16 \times 9/4 $. 

Multiple choice bohr's model of atom structure of atom

What are the values of $n _{1}$ and $n _{2}$ respectively for $H _{\beta}$ line in the Lyman series of hydrogen atomic spectrum?

  1. 3 and 5

  2. 2 and 3

  3. 1 and 3

  4. 2 and 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The $\beta$ line of any series means the second line of that series. Similarly the $\alpha$ line implies the first line and the $\gamma$ line implies the third line of the series.

Multiple choice bohr's model of atom structure of atom

The first emission line of hydrogen atomic spectrum in the Balmer series appears at (R = Rydberg constant):

  1. $\frac{5R}{36}cm^{-1}$
  2. $\frac{3R}{4}cm^{-1}$
  3. $\frac{7R}{144}cm^{-1}$
  4. $\frac{9R}{400}cm^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Use the formula ,
wavenumber =R×(1/n21−1/n22)=R×(1/n12−1/n22).

Multiple choice bohr's model of atom structure of atom

The spectrum of helium is expected to be similar to that of:

  1. $H$
  2. $Li^+$
  3. $Na$
  4. $He^+$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

No. of electrons in$Li^{+}$ & $He = 2$. Therefore the spectrum will be same for both of them with same no. of electrons.

Multiple choice bohr's model of atom structure of atom

The distance between 3rd and 2nd orbits in the hydrogen atom is:

  1. $2.646\times {10}^{-8}cm$
  2. $2.116\times {10}^{-8}cm$
  3. $1.058\times {10}^{-8}cm$
  4. $2.646\times {10}^{-10}cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Solution:

Distance between the 2nd and the 3rd orbits = distance of 3rd orbit – distance of second orbit from the nucleus.

${ d } _{ 3 }=\frac { 0.529\times { n }^{ 2 } }{ z\times { 10 }^{ -10 }m } $

$=0.529\times 9\times { 10 }^{ -10 }m$

${ d } _{ 2 }=0.529\times 4\times { 10 }^{ -10 }m$

${ d } _{ 3 }–{ d } _{ 2 }=0.529\times 5\times { 10 }^{ -10 }m$

  $=2.645\times { 10 }^{ -10 }m$  is the distance between the 2nd the 3rd orbits.

Multiple choice bohr's model of atom structure of atom

The energy of second Bohr orbit of the hydrogen atom is $-328kJ$ ${mol}^{-1}$. Hence the energy of fourth Bohr orbit would be:

  1. $-41kJ$ ${mol}^{-1}$
  2. $-13121kJ$ ${mol}^{-1}$
  3. $-164kJ$ ${mol}^{-1}$
  4. $-82kJ$ ${mol}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
According to the Bohr model, the energy of the orbit is inversely proportional to the square of orbit number:

$E\quad \alpha \quad 1/n²$ -------- (1)

$\dfrac{E _1}{E _2} = \dfrac{n _2}{n _1}$

$\dfrac{-328}{E _2} = \dfrac{4^2}{2^2}$

$E _2 =\dfrac{-328kJ/mol}4$

$=-82kJ/mol.$

Hence, $-82kJ/mol$ is the answer.
Multiple choice bohr's model of atom structure of atom

The shortest $\lambda$ for the Lyman series of hydrogen atom is:


[Given, $R _H=109678 cm^{-1}]$

  1. $911.7A^o$
  2. $700 A^o$
  3. $600 A^o$
  4. $811 A^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
For Lyman series, $n _1=1$

For shortest $\lambda$ of Lyman series; energy difference in two levels showing transition should be maximum i.e $n _2= \infty$

$\dfrac{1}{\lambda}$$=R _H \left[\dfrac{1}{{1}^2}− \dfrac{1}{{(\infty})^2} \right]$

$\dfrac{1}{\lambda}$$=109678$

$\lambda$ $=911.7\times 10^{−8}$ cm

   $=911.7A^0$

Hence, option A is correct.
Multiple choice bohr's model of atom structure of atom

The first emission line in the atomic spectrum of hydrogen in the Balmer Series appears at:

  1. $\dfrac {9R _H}{400}cm^{-1}$
  2. $\dfrac {7R _H}{144}cm^{-1}$
  3. $\dfrac {3R _H}{4}cm^{-1}$
  4. $\dfrac {5R _H}{36}cm^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac {1}{\lambda}=R _HZ^2[\dfrac {1}{n _1^2}-\dfrac {1}{n _2^2}]$
For Balmer Series' first emission line,
$\dfrac {1}{\lambda}=R _H\times 1[\dfrac {1}{2^2}-\dfrac {1}{3^2}]=\dfrac {R _H5}{36}cm^{-1}$

Multiple choice bohr's model of atom structure of atom

Statement I : Wavelength of limiting line of lyman series is less than wavelength of limiting line of Balmer series.
Statement II: Rydberg constant value is same for all elements

  1. Statement I is true, Statement II is also true; Statement is the correct explanation of Statement I

  2. Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I

  3. Statement I is true, Statement II is false

  4. Statement I is false, Statement II is true
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

                              Lyman series                           Balmer series
Limiting line   :         $n=\infty :to:n=1$               $n=\infty :to:n=2$

                                                    $E _l > E _B$                           [limiting case]
                                                    $\lambda _l < \lambda _B$
Rydberg constant represents the limiting value of the highest wavenumber of any photon that can be emitted from an atom.
                                                    $\frac{1}{\lambda _l }=Z^2R        [\lambda _l =limiting :case]$

Multiple choice bohr's model of atom structure of atom

Which are correct for emission spectra of Balmer series in $H$-atom?

  1. $\displaystyle\lambda _{(in nm)}=364.56\left[\frac{n^2 _2}{n^2 _2-n^2 _1}\right]$; where $n _1=2$ and $n _2 > 2$
  2. $\dfrac{1}{\lambda}=R\left[\displaystyle\frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right];$ where $n _1=2$ and $n _2 > 2$; $R=3.29\times 10^{15}H _z$
  3. $\displaystyle\frac{1}{\lambda}=R _H \left[ \frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right ];$ where $n _1=2$ and $n _2 > 2$; $R _H=1.09737\times 10^5cm^{-1}$
  4. $\dfrac{1}{\lambda}=\frac{4c(in msec^{-1})}{364.56\times 10^{-9}}\left[\frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right];$ where $n _1=2$ and $n _2 > 2$;

    $c$ is speed of light.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

From Bohr model, we know that
in a transition,
$\displaystyle\frac{1}{\lambda}=R _H \left[ \frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right ]$ where, $R _H = 109700cm^-$
For Balmer series, electron gets deexcited from 3rd or upper level to second level so $n _1 =2, n _2 >2$.
Also we know that,
$E=hc/\lambda$ so
$\displaystyle v=\frac{4c(in msec^{-1})}{364.56\times 10^{-9}}\left[\frac{1}{n^2 _1}-\frac{1}{n^2 _2}\right];$ where $n _1=2$ and $n _2 > 2$; $c$ is speed of light.

Multiple choice bohr's model of atom structure of atom

An $e^{-}$ of $He^{+}$ makes a transition and emits $6^{th}$ line of Balmer series. Similar wavelength of radiation is absorbed by hydrogen like specie to give $9^{th}$ line of paschen series in its spectrum. The value of Z of the hydrogen like specie is :

  1. 1

  2. 2

  3. 3

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

Balmer $6^{th}$ line (8---2)
Paschen $9^{th}$ line (12---3)
For $\displaystyle He^{+}\frac{1}{\lambda }=R.2^{2}\left ( \frac{1}{2^{2}}-\frac{1}{8^{2}} \right )$

For $He$ atomic number is 2.
$\displaystyle \frac{1}{\lambda }=R\left ( \frac{1}{(2/2)^{2}}-\frac{1}{(8/2)^{2}} \right )$ ....(1)
Fo\displaystyle r single electron species having atomic number 't'
$\displaystyle \frac{1}{\lambda }=Rt^{2}\left ( \frac{1}{3^{2}}-\frac{1}{12^{2}}\right )$
Here t corresponds to atomic number Z of the element.
$\displaystyle \frac{1}{\lambda }=R\left ( \frac{1}{(3/t)^{2}}-\frac{1}{(12/t)^{2}} \right )$ ...(2)
Comparing (1) & (2)
$\Rightarrow \displaystyle \left ( \frac{2}{2} \right )^{2}=\left ( \frac{3}{t} \right )^{2}$
$\Rightarrow Z=3$