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The ratio of maximum to minimum possible radiation energy in Bohr’s hypothetical hydrogen atom is equal to
(A) $2$
(B) $4$
(C) $\dfrac{4}{3}$
(D) $\dfrac{3}{2}$

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Last updated date: 16th Jun 2024
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Answer
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Hint: The ratio of the maximum possible radiation energy to the minimum possible radiation energy in Bohr’s hypothetical hydrogen atom is determined by using the energy of the radiation formula. Then by dividing the maximum energy to the minimum energy, the ratio can be determined.
Formula used:
The energy of the radiation can be given by,
$E = 13.6\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where, $E$ is the energy of the radiation and, ${n_1}$ and ${n_2}$ are the order of the radiation.

Complete step by step solution:
The energy of the radiation can be given by,
$E = 13.6\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)\,.......................\left( 1 \right)$
For the maximum radiation energy, the ${n_1} = 1$ and the ${n_2} = \infty $,
By substituting the ${n_1}$ and ${n_2}$ value in the above equation (1), then the above equation (1) is written as,
$E = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{\infty }} \right)$
Now, the value of the $\dfrac{1}{\infty } = 0$ is substituted in the above equation, then the above equation is written as,
$E = 13.6\left( {1 - 0} \right)$
By subtracting the terms in the above equation, then the above equation is written as,
$E = 13.6\left( 1 \right)$
By multiplying the terms in the above equation, then the above equation is written as,
${E_{\max }} = 13.6\,eV$
For the minimum radiation energy, the ${n_1} = 1$ and the ${n_2} = 2$,
By substituting the ${n_1}$ and ${n_2}$ value in the above equation (1), then the above equation (1) is written as,
$E = 13.6\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)$
By squaring the terms in the above equation, then the above equation is written as,
$E = 13.6\left( {\dfrac{1}{1} - \dfrac{1}{4}} \right)$
By cross multiplying the terms in the above equation, then the above equation is written as,
$E = 13.6\left( {\dfrac{{4 - 1}}{4}} \right)$
By subtracting the terms in the above equation, then the above equation is written as,
${E_{\min }} = 13.6\left( {\dfrac{3}{4}} \right)\,eV$
Now, the ratio of the maximum energy to the minimum energy is given by,
$\dfrac{{{E_{\max }}}}{{{E_{\min }}}} = \dfrac{{13.6}}{{13.6\left( {\dfrac{3}{4}} \right)}}$
By cancelling the same terms in the above equation, then the above equation is written as,
$\dfrac{{{E_{\max }}}}{{{E_{\min }}}} = \dfrac{1}{{\left( {\dfrac{3}{4}} \right)}}$
By rearranging the terms in the above equation, then the above equation is written as,
$\dfrac{{{E_{\max }}}}{{{E_{\min }}}} = \dfrac{4}{3}$

Hence, the option (C) is the correct answer.

Note: In the question it is asked that the ratio of the maximum possible radiation energy to the minimum possible radiation energy in Bohr’s hypothetical hydrogen atom, so the maximum energy is divided by the minimum energy of the radiation energy.