In an H-like atom,when an electron transits f | vedclass

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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

Vedclass · Accepted Answer

$486$

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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.Since $R$ and $Z$ are constant for a given atom,we have $\frac{1}{\lambda} \propto \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the first transition ($n=5$ to $n=2$): $\frac{1}{434} = k \left( \frac{1}{2^2} - \frac{1}{5^2} ight) = k \left( \frac{1}{4} - \frac{1}{25} ight) = k \left( \frac{21}{100} ight)$.For the second transition ($n=4$ to $n=2$): $\frac{1}{x} = k \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = k \left( \frac{1}{4} - \frac{1}{16} ight) = k \left( \frac{3}{16} ight)$.Dividing the two equations: $\frac{x}{434} = \frac{21/100}{3/16} = \frac{21}{100} 	imes \frac{16}{3} = \frac{7 	imes 16}{100} = \frac{112}{100} = 1.12$.Thus,$x = 434 	imes 1.12 = 486.08 \, nm \approx 486 \, nm$.

In an $H$-like atom,when an electron transits from energy state $n=5$ to $n=2$,a photon of wavelength $434 \, nm$ is emitted. What will be the wavelength of the photon emitted when the transition occurs from energy state $n=4$ to $n=2$?

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