Tag: structure of atom

Questions Related to structure of atom

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.

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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

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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