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QJEEM23Ph74cf: JEE Main 2023 Physics - Class 12 JEEM23Ph74cf Solution: For hydrogen atom, and are the wavelengths corresponding to the …

For hydrogen atom, λ1\lambda_{1} and λ2\lambda_{2} are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of λ1\lambda_{1} and λ2\lambda_{2} is x32\frac{x}{32} . The value of xx is __________.

JEE Main 2023 (Online) 31st January Morning Shift Physics - Atoms and Nuclei Question 85 English

Step-by-Step Verified Solution

1λ=RZ2[1n12−1n22]\frac{1}{\lambda}=\mathrm{RZ}^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]

amp;1λ1=RZ2[112−132]=89RZ2........(1)amp;1λ2=RZ2[112−122]=34RZ2........(2)\begin{aligned}& \frac{1}{\lambda_{1}}=\mathrm{RZ}^{2}\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]=\frac{8}{9} \mathrm{RZ}^{2} ........(1)\\\\ & \frac{1}{\lambda_{2}}=\mathrm{RZ}^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=\frac{3}{4} \mathrm{RZ}^{2} ........(2)\end{aligned}

1/2⇒λ2λ1=89×43=3227λ1λ2=2732\begin{gathered}1 / 2 \Rightarrow \frac{\lambda_{2}}{\lambda_{1}}=\frac{8}{9} \times \frac{4}{3}=\frac{32}{27} \\\\\frac{\lambda_{1}}{\lambda_{2}}=\frac{27}{32}\end{gathered}

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