Hydrogen (H),deuterium (D),singly ionized hel | vedclass

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(A) For a hydrogen-like atom,the wavelength $\lambda$ of the emitted radiation for a transition from $n_2$ to $n_1$ is given by the Rydberg formula: $

Vedclass · Accepted Answer

$\lambda_1 = \lambda_2 = 4 \lambda_3 = 9 \lambda_4$

Vedclass · Answer

(A) For a hydrogen-like atom,the wavelength $\lambda$ of the emitted radiation for a transition from $n_2$ to $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the transition $n = 2$ to $n = 1$,we have $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R Z^2 \left( 1 - \frac{1}{4} ight) = \frac{3R Z^2}{4}$.Thus,$\lambda = \frac{4}{3RZ^2}$,which implies $\lambda Z^2 = 	ext{constant}$.Given the atomic numbers: $Z_1 = 1$ (for $H$),$Z_2 = 1$ (for $D$),$Z_3 = 2$ (for $He^+$),and $Z_4 = 3$ (for $Li^{++}$).Substituting these values: $\lambda_1 (1)^2 = \lambda_2 (1)^2 = \lambda_3 (2)^2 = \lambda_4 (3)^2$.This simplifies to $\lambda_1 = \lambda_2 = 4 \lambda_3 = 9 \lambda_4$.

Hydrogen $(H)$,deuterium $(D)$,singly ionized helium $(He^+)$ and doubly ionized lithium $(Li^{++})$ all have one electron around the nucleus. Consider the $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ respectively.

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