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QJEEM23Ph04a2: JEE Main 2023 Physics - Class 12 JEEM23Ph04a2 Solution: The wavelength of the radiation emitted is when an electron…

The wavelength of the radiation emitted is λ0\lambda_0 when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will 20xλ0\frac{20}{x}\lambda_0 . The value of xx is _____________.

Step-by-Step Verified Solution

Second excited state →\to first excited state , n=3n=3 to n=2n=2

1λ0=R(14−19)=(5R36)\frac{1}{\lambda_{0}}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\left(\frac{5 R}{36}\right)

Third excited state →\to second orbit , n=4n=4 to n=2n=2

1λ=R(14−116)=(316R)\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{16}\right)=\left(\frac{3}{16} R\right)

Taking ratio of (1) and (2)

λλ0=536×163=(2027)\frac{\lambda}{\lambda_{0}}=\frac{5}{36} \times \frac{16}{3}=\left(\frac{20}{27}\right)

λ=2027λ0\lambda=\frac{20}{27} \lambda_{0}

x=27x=27

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