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(C) The energy difference $\Delta E$ for a transition in a hydrogen atom is given by $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

Vedclass · Accepted Answer

$\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$

Vedclass · Answer

(C) The energy difference $\Delta E$ for a transition in a hydrogen atom is given by $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) ext{ eV}$.The wavelength $\lambda$ is inversely proportional to the energy difference: $\lambda = \frac{hc}{\Delta E}$.Thus,a larger $\Delta E$ corresponds to a smaller $\lambda$.Calculating relative energy differences:$(a) \ \Delta E_1 \propto (1/1^2 - 1/3^2) = 1 - 0.111 = 0.889$$(b) \ \Delta E_3 \propto (1/10^2 - 1/12^2) = 0.01 - 0.0069 = 0.0031$$(c) \ \Delta E_2 \propto (1/3^2 - 1/5^2) = 0.111 - 0.04 = 0.071$$(d) \ \Delta E_4 \propto (1/20^2 - 1/22^2) = 0.0025 - 0.00206 = 0.00044$Comparing energy differences: $\Delta E_1 > \Delta E_2 > \Delta E_3 > \Delta E_4$.Therefore,the wavelengths follow the order: $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$.

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