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# Monochromatic radiation of wavelength λ is incident on a hydrogen sample containing ground state. Hydrogen atoms absorb the light and subsequently emit radiations of ten different wavelengths. The value of λ is.A. 95nmB.103nmC.73nmD.88nm

Last updated date: 20th Jun 2024
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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
$=-13.6\times \dfrac{{{Z}^{2}}}{{{n}^{2}}}$
$\Delta E=\dfrac{hc}{\lambda }$

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.