Answer
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Hint: Firstly we will see that the atom is absorbing emitted radiation and making transition to higher state from ground state. So it will find its higher state. We must know that for transition some energy is required we will find $\Delta E$. Afterward, we will find the required wavelength by using the formula of energy.
Formula used:
Total no of lines
$=\dfrac{n\left( n-1 \right)}{2}$
According to Bohr model
Energy in n state
$=-13.6\times \dfrac{{{Z}^{2}}}{{{n}^{2}}}$
$\Delta E=\dfrac{hc}{\lambda }$
Complete step by step answer:
Firstly we know that the total no of lines emitted are 10 so we will firstly find its highest state.
We know the formula
Total no of lines
$=\dfrac{n\left( n-1 \right)}{2}$
$\begin{align}
& 10=\dfrac{n\left( n-1 \right)}{2} \\
& 20={{n}^{2}}-n \\
& n=5 \\
\end{align}$
So the atom has attained fifth higher state.
Now for transition in states atom require some energy which is given by
$\Delta {{E}_{c}}={{E}_{{{n}_{2}}}}-{{E}_{{{n}_{1}}}}$
According to Bohr’s atomic model
${{E}_{n}}=-13.6\times \dfrac{{{Z}^{2}}}{n}$
So for hydrogen sample Z=1
\[\begin{align}
& \Delta E=13.6\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right] \\
& \\
\end{align}\]
Atom is going from its ground state to fifth state.
\[\begin{align}
& \Delta E=13.6\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{5}^{2}}} \right] \\
& \Delta E=\dfrac{hc}{\lambda } \\
& \dfrac{hc}{\lambda }=13.6\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{5}^{2}}} \right]\text{e}\text{.V} \\
& \dfrac{1240}{\lambda }=13.6\times \dfrac{24}{25}\approx 95\text{nm} \\
\end{align}\]
So the correct option is A.
Additional Information:
We must know that the energy of light emitted and absorbed is just equal to the difference between the energy of orbitals. The energy of orbits are always quantized. Bohr also states that energy of the orbital is directly proportional to distance of the nucleus from the orbital.
We should also know that the atom absorbs or emits light in discrete packets called photons, and which carries a definite amount of energy.
Note:
According to Bohr’s atomic model, electrons travel in circular orbit around the nucleus which is well defined. It also states that electrons can jump from one orbit to another just by emitting or absorbing energy. In the hydrogen atom, electrons are travelling in predefined atomic orbitals. Radiation is emitted when electrons fall from higher state to lower state and absorbed when electron jumps to higher state.
Formula used:
Total no of lines
$=\dfrac{n\left( n-1 \right)}{2}$
According to Bohr model
Energy in n state
$=-13.6\times \dfrac{{{Z}^{2}}}{{{n}^{2}}}$
$\Delta E=\dfrac{hc}{\lambda }$
Complete step by step answer:
Firstly we know that the total no of lines emitted are 10 so we will firstly find its highest state.
We know the formula
Total no of lines
$=\dfrac{n\left( n-1 \right)}{2}$
$\begin{align}
& 10=\dfrac{n\left( n-1 \right)}{2} \\
& 20={{n}^{2}}-n \\
& n=5 \\
\end{align}$
So the atom has attained fifth higher state.
Now for transition in states atom require some energy which is given by
$\Delta {{E}_{c}}={{E}_{{{n}_{2}}}}-{{E}_{{{n}_{1}}}}$
According to Bohr’s atomic model
${{E}_{n}}=-13.6\times \dfrac{{{Z}^{2}}}{n}$
So for hydrogen sample Z=1
\[\begin{align}
& \Delta E=13.6\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right] \\
& \\
\end{align}\]
Atom is going from its ground state to fifth state.
\[\begin{align}
& \Delta E=13.6\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{5}^{2}}} \right] \\
& \Delta E=\dfrac{hc}{\lambda } \\
& \dfrac{hc}{\lambda }=13.6\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{5}^{2}}} \right]\text{e}\text{.V} \\
& \dfrac{1240}{\lambda }=13.6\times \dfrac{24}{25}\approx 95\text{nm} \\
\end{align}\]
So the correct option is A.
Additional Information:
We must know that the energy of light emitted and absorbed is just equal to the difference between the energy of orbitals. The energy of orbits are always quantized. Bohr also states that energy of the orbital is directly proportional to distance of the nucleus from the orbital.
We should also know that the atom absorbs or emits light in discrete packets called photons, and which carries a definite amount of energy.
Note:
According to Bohr’s atomic model, electrons travel in circular orbit around the nucleus which is well defined. It also states that electrons can jump from one orbit to another just by emitting or absorbing energy. In the hydrogen atom, electrons are travelling in predefined atomic orbitals. Radiation is emitted when electrons fall from higher state to lower state and absorbed when electron jumps to higher state.
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