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This is one of our many quizzes on Bohr model. Click on this to access more such competitive quizzes on Structure Of An Atom.

Bohr model Quiz 2

chemistry

From the below given atomic models, which has properly explained the stability of an atom and has described the line spectrum of hydrogen atom but has failed to state the wave nature of an electron? 

  Bohr’s model had some drawbacks which were then described by the dual nature of matter i.e. considering wave spectrum of electron as well.

Bohr’s model explained the stability of an atom and has described the line spectrum of hydrogen atom but has failed to state the wave nature of an electron.

We are familiar with the spectral lines generated by the electrons when it jumps from one orbit to another. So, tell us which transition corresponds to I. third line in Balmer series and II. second line in the Paschen series of the hydrogen spectrum?

 The diagram can help to solve this better; but if we want to solve orally, then do note that Balmer has 3 spectrums and Paschen series has 2 spectrums in total.

According to the diagram,<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/1680aed3-c4b6-4d1a-942a-ef0053e811127935304579560117511.png"/> The third line in the Balmer series corresponds to the transition from n = 3 to n = 2. The second line in the Paschen series corresponds to the transition from n = 4 to n = 3.

Considering the diagram of spectral analysis of the H – atom, calculate the total spectral lines in the Lyman series i.e. from level 5 to 1 level.<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/acc0df97-edf9-4245-9024-29b8076942a83173253986748252057.png"/>

 To solve this problem, we must know the basic formula which has been derived from the spectral diagram i.e.$N=\dfrac{\left( {{n}_{2}}-{{n}_{1}} \right)\left( {{n}_{2}}-{{n}_{1}}+1 \right)}{2}$ where, N is the number of emission lines when an electron jumps from ${{n}_{2}}$ level to ${{n}_{1}}$ level.

 $ \text{The number of emission lines (N) when an electron jumps from }{{n}_{2}}\text{ level to }{{n}_{1}}\text{ level is given by;}\\ N=\dfrac{\left( {{n}_{2}}-{{n}_{1}} \right)\left( {{n}_{2}}-{{n}_{1}}+1 \right)}{2}\\ \text{Here, }{{n}_{1}}=1\text{ and }{{n}_{2}}=5\text{ then;}\\ N=\dfrac{\left( {{n}_{2}}-{{n}_{1}} \right)\left( {{n}_{2}}-{{n}_{1}}+1 \right)}{2} \\ N=\dfrac{\left( 5-1 \right)\left( 5-1+1 \right)}{2}=10 \\ \text{OR,}\\ \text{We can also calculate these lines by one simple method;}\\ 5\to 4,5\to 3,5\to 2,5\to 1\Rightarrow 4 \\ 4\to 3,4\to 2,4\to 1\Rightarrow 3 \\ 3\to 2,3\to 1\Rightarrow 2 \\ 2\to 1\Rightarrow 1 \\ \text{Total }=4+3+2+1=10lines $

 We have known the energy associated with the first orbit in the hydrogen atom i.e. $2.17\times {{10}^{-18}}Jato{{m}^{-1}}$ . What will be energy associated with the fourth orbit? Also, calculate the radius of the fourth orbit.

 This can be easily solved by the equations given by the Bohr’s atomic model for energy i.e. ${{E}_{n}}=-\dfrac{{{E}_{1}}}{{{n}^{2}}}Jato{{m}^{-1}}$ and radius i.e. ${{r}_{n}}={{r}_{1}}{{n}^{2}}nm$ of any orbit.

 $ \text{We know that,}\\ \text{The energy associated with }{{n}^{th}}\text{ orbit in hydrogen atom is,}\\ {{E}_{n}}=-\dfrac{{{E}_{1}}}{{{n}^{2}}}Jato{{m}^{-1}}\text{ and }\\ \text{Radius of }{{n}^{th}}\text{ orbit is,}\\ {{r}_{n}}={{r}_{1}}{{n}^{2}}nm \\.\text{Now, we know that }\\ {{E}_{1}}=2.17\times {{10}^{-18}}Jato{{m}^{-1}}\text{ and }{{r}_{1}}=0.0529nm.\\ \text{So, for n = 4;}\\ {{E}_{4}}=-\dfrac{2.17\times {{10}^{-18}}}{{{4}^{2}}}=-\dfrac{2.17\times {{10}^{-18}}}{16}=-1.36\times {{10}^{-19}}Jato{{m}^{-1}}\text{and}\\ {{r}_{4}}=0.0529\times {{4}^{2}}=0.0529\times 16=0.8464nm$

We have studies Bohr’s theory; from that studies we can say that the electronic energy of hydrogen atom is the ${{n}^{th}}$ Bohr orbit is given as ${{E}_{n}}=-\dfrac{21.76\times {{10}^{-19}}}{{{n}^{2}}}Jato{{m}^{-1}}$ Calculate the longest wavelength of light that will be needed to remove an electron from the fourth orbit of $H{{e}^{+}}$ ion.

 The energy formulations can help us solve this problem i.e. $\Delta E={{E}_{\infty }}-{{E}_{n}}=\dfrac{hc}{\lambda }$ .

 We have known that energy relations are same for hydrogen atom and hydrogen like atoms with a variation as;$ \\ \text{Energy of electron in }{{n}^{th}}\text{ orbit in }H{{e}^{+}} ion;\\ {{E}_{n}}=-\dfrac{21.76\times {{10}^{-19}}{{Z}^{2}}}{{{n}^{2}}}Jato{{m}^{-1}}\\ \text{where, Z is the atomic number i.e. 2 for }H{{e}^{+}}.\\ \text{Energy of }H{{e}^{+}}\text{ ion in the fourth orbit is given as;}\\ {{E}_{4}}=-\dfrac{21.76\times {{10}^{-19}}\times {{2}^{2}}}{{{4}^{2}}} \\ {{E}_{4}}=-\dfrac{21.76\times {{10}^{-19}}\times 4}{16} \\ \text{Energy required to remove the electron from fourth orbit;}\\ \Delta E={{E}_{\infty }}-{{E}_{4}} \\ \Delta E=0-\left( -\dfrac{21.76\times {{10}^{-19}}\times 4}{16} \right) \\ \Delta E=5.44\times {{10}^{-19}}J \\ \text{Now,}\\ \Delta E=\dfrac{hc}{\lambda }\\ \text{where, h is the planck's constant}\\ \lambda =\dfrac{hc}{\Delta E}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5.44\times {{10}^{-19}}}=3.654\times {{10}^{-7}}m \\ \lambda =365.4nm $