Energy levels A, B, C of a certain atom corre | vedclass

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(B) The energy of a photon emitted during a transition between two energy levels $E_i$ and $E_f$ is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda

Vedclass · Accepted Answer

$\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}$

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(B) The energy of a photon emitted during a transition between two energy levels $E_i$ and $E_f$ is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda}$.From the given energy level diagram,the transition from $C$ to $A$ is the sum of the transitions from $C$ to $B$ and $B$ to $A$.Therefore,the energy difference is:$E_C - E_A = (E_C - E_B) + (E_B - E_A)$Substituting the energy-wavelength relation $\Delta E = \frac{hc}{\lambda}$:$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$Dividing both sides by $hc$:$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$$\frac{1}{\lambda_3} = \frac{\lambda_2 + \lambda_1}{\lambda_1 \lambda_2}$Taking the reciprocal,we get:$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$

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