Tag: structure of atom

Questions Related to 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)

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

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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. $He^+$

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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.
                                                    $\frac{1}{\lambda _l }=Z^2R        [\lambda _l =limiting :case]$