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

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(A) The energy of a photon emitted during a transition from $n_2$ to $n_1$ is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The energy of a photon emitted during a transition from $n_2$ to $n_1$ is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Since the transition $n=2$ to $n=1$ is the same for all,$\Delta E \propto Z^2$.We know that $\Delta E = \frac{hc}{\lambda}$,therefore $\frac{hc}{\lambda} \propto Z^2$,which implies $\lambda Z^2 = 	ext{constant}$.For Hydrogen $(Z=1)$,$\lambda_1 Z_1^2 = \lambda_1 (1)^2 = \lambda_1$.For Deuterium $(Z=1)$,$\lambda_2 Z_2^2 = \lambda_2 (1)^2 = \lambda_2$.For Helium $(Z=2)$,$\lambda_3 Z_3^2 = \lambda_3 (2)^2 = 4\lambda_3$.For Lithium $(Z=3)$,$\lambda_4 Z_4^2 = \lambda_4 (3)^2 = 9\lambda_4$.Equating these,we get $\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

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