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Answer :
Final answer:
The calculated wavelength of radiation emitted in the Lyman series when an electron falls from the fourth energy level to the first in a hydrogen atom is approximately 97.0 nm, which is closest to option (c) 97.3 nm. The correct option is c.
Explanation:
The question involves calculating the wavelength of radiation emitted in the Lyman series when an electron falls from the fourth energy level (n=4) to the first energy level (n=1) in a hydrogen atom. To find the wavelength, we use the Rydberg formula for hydrogen:
\( \frac{1}{\lambda} = R ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ) \)
where \( \lambda \) is the wavelength of the emitted photon, R is the Rydberg constant (which is given as 1.1 \( \times \) 107 m-1), and n1 and n2 are the principal quantum numbers of the lower and upper energy levels, respectively. In this case, the electron drops from n2=4 to n1=1.
By substituting n1=1 and n2=4 into the Rydberg formula, we get:
\( \frac{1}{\lambda} = 1.1 \( \times \) 107 m-1 ( \frac{1}{1^2} - \frac{1}{4^2} ) = 1.1 \( \times \) 107 m-1 \( \frac{15}{16} \) \)
\( \frac{1}{\lambda} = 1.03125 \( \times \) 107 m-1 \)
Therefore, the wavelength \( \lambda \) is:
\( \lambda = \frac{1}{1.03125 \( \times \) 107 m-1} = 9.695 \( \times \) 10-8 m = 96.95 nm \)
The correct wavelength of radiation emitted is then approximately 97.0 nm when rounding to three significant figures, which corresponds to option (c) 97.3 nm. The correct option is c.
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