Hydrogen (_1H^1),Deuterium (_1H^2),singly ion | vedclass

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(B) The wavelength $\lambda$ of emitted radiation for a hydrogen-like atom is given by the Rydberg formula:$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The wavelength $\lambda$ of emitted radiation for a hydrogen-like atom is given by the Rydberg formula:$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$Since the transition is from $n_2 = 2$ to $n_1 = 1$ for all ions,the term in the bracket is constant: $\left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = \frac{3}{4}$.Thus,$\lambda \propto \frac{1}{Z^2}$.For Hydrogen $(Z=1)$,$\lambda_1 \propto \frac{1}{1^2} = 1$.For Deuterium $(Z=1)$,$\lambda_2 \propto \frac{1}{1^2} = 1$.For Helium $(Z=2)$,$\lambda_3 \propto \frac{1}{2^2} = \frac{1}{4}$.For Lithium $(Z=3)$,$\lambda_4 \propto \frac{1}{3^2} = \frac{1}{9}$.Comparing these,we get $\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

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