• Bohr's model of atom - class-X
Back
0

Bohr's model of atom - class-X

Description: Bohr's model of atom
Number of Questions: 16
Created by:
Tags: structure of atom
Start the Quiz
Attempted 0/16 Correct 0 Score 0
‹
›

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)

bohr's model of atom structure of atom

  1. $\dfrac{5hcR}{36}$

  2. $\dfrac{4hcR}{3}$

  3. $\dfrac{3hcR}{4}$

  4. $\dfrac{7hcR}{144}$

Show answer
Correct Option: C
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$=2$.

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:

bohr's model of atom structure of atom

  1. 21/16 x 9/4

  2. 25/16 x 9/4

  3. 21/25 x 9/4

  4. 16/25 x 9/4

Show answer
Correct Option: A
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 $. 

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

bohr's model of atom structure of atom

  1. 3 and 5

  2. 2 and 3

  3. 1 and 3

  4. 2 and 4

Show answer
Correct Option: C
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.

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

bohr's model of atom structure of atom

  1. $\frac{5R}{36}cm^{-1}$

  2. $\frac{3R}{4}cm^{-1}$

  3. $\frac{7R}{144}cm^{-1}$

  4. $\frac{9R}{400}cm^{-1}$

Show answer
Correct Option: A
Explanation:

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

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

bohr's model of atom structure of atom

  1. $H$

  2. $Li^+$

  3. $Na$

  4. $He^+$

Show answer
Correct Option: B
Explanation:

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

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

bohr's model of atom structure of atom

  1. $2.646\times {10}^{-8}cm$

  2. $2.116\times {10}^{-8}cm$

  3. $1.058\times {10}^{-8}cm$

  4. $2.646\times {10}^{-10}cm$

Show answer
Correct Option: A
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.

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

bohr's model of atom structure of atom

  1. $-41kJ$ ${mol}^{-1}$

  2. $-13121kJ$ ${mol}^{-1}$

  3. $-164kJ$ ${mol}^{-1}$

  4. $-82kJ$ ${mol}^{-1}$

Show answer
Correct Option: D
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.

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


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

bohr's model of atom structure of atom

  1. $911.7A^o$

  2. $700 A^o$

  3. $600 A^o$

  4. $811 A^o$

Show answer
Correct Option: A
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.

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

bohr's model of atom structure of atom

  1. $\dfrac {9R _H}{400}cm^{-1}$

  2. $\dfrac {7R _H}{144}cm^{-1}$

  3. $\dfrac {3R _H}{4}cm^{-1}$

  4. $\dfrac {5R _H}{36}cm^{-1}$

Show answer
Correct Option: D
Explanation:

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

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

bohr's model of atom structure of atom

  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

Show answer
Correct Option: C
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.
                                                    $\frac{1}{\lambda _l }=Z^2R        [\lambda _l =limiting :case]$

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

bohr's model of atom structure of 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.

Show answer
Correct Option: A,B,C,D
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.

In which transition one quantum of energy is emitted?

bohr's model of atom structure of atom

  1. $\displaystyle n=4\rightarrow n=2$

  2. $\displaystyle n=3\rightarrow n=1$

  3. $\displaystyle n=4\rightarrow n=1$

  4. All of them

Show answer
Correct Option: D
Explanation:

In all the three transitions $\displaystyle n=4\rightarrow n=2$, $\displaystyle n=3\rightarrow n=1$ and $\displaystyle n=4\rightarrow n=1$, one quantum of energy is emitted. Whenever an electron jumps from higher energy level to lower energy level, a quantum of energy is emitted.

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 :

bohr's model of atom structure of atom

  1. 1

  2. 2

  3. 3

  4. 4

Show answer
Correct Option: C
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)
$\Rightarrow \displaystyle \left ( \frac{2}{2} \right )^{2}=\left ( \frac{3}{t} \right )^{2}$
$\Rightarrow Z=3$

In a mixture of $H-He^{+}$ gas, H atom and $He^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer its excitation energy to $He^{+}$ ions by collision.
If each hydrogen atom in the ground state is excited by absorbing photons of energy 8.4 eV, 12.09 eV, of energy, then assuming the Bohr model of an atom is applicable the number of spectral lines emitted is equal to:

bohr's model of atom structure of atom

  1. 5

  2. 2

  3. 3

  4. 4

Show answer
Correct Option: C
Explanation:

Energy of electron in $n^{th}$ shell:


$E _n=  \dfrac{–13.12\ Z^2}{n^2} eV$


$n= 1: -13.6 eV$
$n= 2: -3.4 eV$
$n= 3: -1.51 eV$
$n=4: -0.85 eV$

$E(n=3)- E(n=1)= 12.09 eV$

So if $H$ atoms are excited by $8.4eV$ and $12.09 eV$ then the electrons will reach to $n=3$ shell. Then the no. of emitted spectral lines will be equal to $3$.


Hence, the correct option is $(C)$.

The emission spectrum of hydrogen is found to satisfy the expression for the energy change $\triangle E$ (in joules) such that $\triangle E = 2.18\times 18^{-18}(\frac{1}{n _1^2}-\frac{1}{n _2^2})J$ where $n _1$= 1, 2, 3, .......and $n _2$ = 2, 3, 4. The spectral lines corresponds to Paschen series if :

bohr's model of atom structure of atom

  1. $n _1 = 1$ and $ n _2 = 2, 3, 4$

  2. $n _1 = 3$ and $ n _2 = 4, 5, 6$

  3. $n _1 = 1$ and $ n _2 = 3, 4, 5$

  4. $n _1 = 2$ and $ n _2 = 3, 4, 5$

Show answer
Correct Option: B
Explanation:

In the emission spectra of hydrogen atom, Paschen series is the one where the transition from higher energy states to third energy state takes place.

i.e. $n _f=n _1=3$ and $n _i=n _2>3$
option B

What would be the wavelength and name of series respectively for the emission transition for H-atom if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm?

bohr's model of atom structure of atom

  1. 434 nm, Balmer

  2. 434 pm, Paschen

  3. 545 pm, Pfund

  4. 600 nm, Lyman

Show answer
Correct Option: A
Explanation:
The radii of the $n^{th}$ stationary state for a hydrogen-like specie is expressed as :

$r _n = \cfrac{n^2 a _0}{Z}$
where $Z=$atomic number and $a _0=52.9\ pm$ radius of Bohr orbit.

For hydrogen atom, Z=1
Given that transition is from orbit radius = 1.3225 nm to 211.6 pm 
Orbit with radius = 1.3225 nm=1322.5 pm

$r _n=52.9 \times n^2=1322.5$

$n^2=25$ or $n=n _i=5$ 

Similarly for Orbit with radius = 211.6 pm

$r _n=52.9 \times n^2=211.6$

$n^2=4$ or $n=n _f=2$

thus transition is from $n _i=5$ to $n _f=2$

Transition energy from $n _i\ to\ n _f$ is given as:
$\frac{1}{\lambda}=R _H[\cfrac{1}{n _f^2}-\cfrac{1}{n _i^2}]$

where $R _H=109677cm^{-1}$ and $n _i=5,n _f=2$

$\cfrac{1}{\lambda}=109677[\cfrac{1}{2^2}-\cfrac{1}{5^2}]\ cm^{-1}$

$\cfrac{1}{\lambda}=109677[\cfrac{1}{4}-\cfrac{1}{25}]\ cm^{-1}$

$\cfrac{1}{\lambda}=109677 \times 0.21\ cm^{-1}$

$\cfrac{1}{\lambda}=23032.17\ cm^{-1}$

$\lambda=4.342\times 10^{-5} cm=434.2\ nm$
Since transition is from higher energy state to n=2, it belongs to Balmer series and wavelength of the transition is 434 nm

- Hide questions
ADVERTISEMENT

Find more quizzes from top tags

general knowledge
3615 Quizzes
119858 Questions
 technology
307 Quizzes
36705 Questions
 sports
963 Quizzes
19148 Questions
 history
1187 Quizzes
13475 Questions
 physics
598 Quizzes
13441 Questions
 biology
1160 Quizzes
12174 Questions
 science & technology
551 Quizzes
11015 Questions
 chemistry
490 Quizzes
9892 Questions
 maths
518 Quizzes
9324 Questions
 softskills
0 Quizzes
7131 Questions