
If the energy of a hydrogen atom in \[{{n}^{th}}\] orbit is ${{E}_{n}}$ then energy in the ${{n}^{th}}$ orbit of a singly ionized helium atom will be ?
Answer
521.1k+ views
Hint:With the help of formula of energy of electron in \[{{n}^{th}}\]orbit, try to find the relation of energy with z(atomic no), n(orbit number) and different variable parameters. Put the values of required parameters in the obtained relation. In this way, you will be able to find the result.
Complete step by step answer:
We know the formula for finding energy of electron in \[{{n}^{th}}\]orbit,
$E=\dfrac{-m{{z}^{2}}{{e}^{4}}}{8{{\varepsilon }_{\circ }}{{h}^{2}}}\cdot \dfrac{1}{{{n}^{^{2}}}}$
Here you can see $m,\,h,\,{{\varepsilon }_{\circ }},\,e$ are constants in the above equation.
So, we can find the relation that $E\,\alpha \,\dfrac{{{z}^{2}}}{{{n}^{2}}}$.
Now, lets’ focus on what we have been given in the question
Given, energy of hydrogen in \[{{n}^{th}}\] orbit = ${{E}_{n}}$
We know, atomic number of hydrogen, ${{Z}_{H}}=1$
We know, atomic number of helium, ${{Z}_{He}}=1$
Here what you can do is to find energy of both hydrogen and helium separately and then find their relation using the relation that we have found after removing the constants $E\,\alpha \,\dfrac{{{z}^{2}}}{{{n}^{2}}}$. The second way is better and less time consuming and we will be doing that for every such question. Now,
$\dfrac{{{E}_{H}}}{{{E}_{He}}}=\dfrac{{{Z}_{H}}^{2}}{{{Z}_{He}}^{2}} \\
\Rightarrow \dfrac{{{E}_{H}}}{{{E}_{He}}}=\dfrac{1}{4} \\
\Rightarrow {{E}_{He}}=4{{E}_{H}} \\ $
Now, we have been given energy of hydrogen in \[{{n}^{th}}\] orbit (i.e.${{E}_{H}}$) =${{E}_{n}}$ , put this in the above findings.
${{E}_{He}}=4{{E}_{H}} \\
\therefore {{E}_{He}}=4{{E}_{n}} \\ $
Hence we find the required energy in the ${{n}^{th}}$ orbit of a singly ionized helium atom = $4{{E}_{n}}$.
Note:In such types of questions, focus on what we have to find. Then try to find the relation from the formula (relation means how is that proportional to the variables). For doing that, simply ignore the constant part from formula and find the proportionality relation.
Complete step by step answer:
We know the formula for finding energy of electron in \[{{n}^{th}}\]orbit,
$E=\dfrac{-m{{z}^{2}}{{e}^{4}}}{8{{\varepsilon }_{\circ }}{{h}^{2}}}\cdot \dfrac{1}{{{n}^{^{2}}}}$
Here you can see $m,\,h,\,{{\varepsilon }_{\circ }},\,e$ are constants in the above equation.
So, we can find the relation that $E\,\alpha \,\dfrac{{{z}^{2}}}{{{n}^{2}}}$.
Now, lets’ focus on what we have been given in the question
Given, energy of hydrogen in \[{{n}^{th}}\] orbit = ${{E}_{n}}$
We know, atomic number of hydrogen, ${{Z}_{H}}=1$
We know, atomic number of helium, ${{Z}_{He}}=1$
Here what you can do is to find energy of both hydrogen and helium separately and then find their relation using the relation that we have found after removing the constants $E\,\alpha \,\dfrac{{{z}^{2}}}{{{n}^{2}}}$. The second way is better and less time consuming and we will be doing that for every such question. Now,
$\dfrac{{{E}_{H}}}{{{E}_{He}}}=\dfrac{{{Z}_{H}}^{2}}{{{Z}_{He}}^{2}} \\
\Rightarrow \dfrac{{{E}_{H}}}{{{E}_{He}}}=\dfrac{1}{4} \\
\Rightarrow {{E}_{He}}=4{{E}_{H}} \\ $
Now, we have been given energy of hydrogen in \[{{n}^{th}}\] orbit (i.e.${{E}_{H}}$) =${{E}_{n}}$ , put this in the above findings.
${{E}_{He}}=4{{E}_{H}} \\
\therefore {{E}_{He}}=4{{E}_{n}} \\ $
Hence we find the required energy in the ${{n}^{th}}$ orbit of a singly ionized helium atom = $4{{E}_{n}}$.
Note:In such types of questions, focus on what we have to find. Then try to find the relation from the formula (relation means how is that proportional to the variables). For doing that, simply ignore the constant part from formula and find the proportionality relation.
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