Tag: structure of atom

Quizzes (9) Questions (16)

Questions Related to structure of atom

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