Tag: bohr's model of atom

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Questions Related to bohr's model of atom

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

In which transition one quantum of energy is emitted?

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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)
$\Rightarrow \displaystyle \left ( \frac{2}{2} \right )^{2}=\left ( \frac{3}{t} \right )^{2}$
$\Rightarrow Z=3$

Multiple choice bohr's model of atom structure of atom

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:

  1. 5

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $(C)$.

Multiple choice bohr's model of atom structure of atom

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 :

  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$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

Multiple choice bohr's model of atom structure of atom

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?

  1. 434 nm, Balmer

  2. 434 pm, Paschen

  3. 545 pm, Pfund

  4. 600 nm, Lyman
Reveal answer Fill a bubble to check yourself
A Correct answer
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