Atoms & Nuclei

JEE Main level, numerical answer, medium

12 of 239

Question 12 of 239

xx+4\dfrac{x}{x+4} is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its
(i) third permitted energy level to the second level and
(ii) the highest permitted energy level to the second permitted level.
The value of xx will be

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Answer: 5

Energy of a photon released in transition of electron in hydrogen atom is given by,
E=13.6(1n22−1n12)eVE=13.6\left(\dfrac{1}{{{n}_{2}}^{2}}-\dfrac{1}{{{n}_{1}}^{2}}\right)eV
⇒E1E2=13.6(14−19)13.6(14−1∞)=xx+4⇒14−1914=xx+4\Rightarrow \dfrac{{E}_{1}}{{E}_{2}}=\dfrac{13.6\left(\dfrac{1}{4}-\dfrac{1}{9}\right)}{13.6\left(\dfrac{1}{4}-\dfrac{1}{\infty }\right)}=\dfrac{x}{x+4} \Rightarrow \dfrac{\dfrac{1}{4}-\dfrac{1}{9}}{\dfrac{1}{4}}=\dfrac{x}{x+4}
⇒59=xx+4\Rightarrow \dfrac{5}{9}=\dfrac{x}{x+4}
⇒5x+20=9x\Rightarrow 5x+20=9x
⇒x=5\Rightarrow x=5

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