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(A) When the electron de-excites from $3E$ to $E$,the energy difference is $\Delta E_1 = 3E - E = 2E$.The energy of the emitted photon is given by $\f

Vedclass · Accepted Answer

$3\lambda$

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(A) When the electron de-excites from $3E$ to $E$,the energy difference is $\Delta E_1 = 3E - E = 2E$.The energy of the emitted photon is given by $\frac{hc}{\lambda} = 2E$ ... $(i)$When the electron de-excites from $\frac{5E}{3}$ to $E$,the energy difference is $\Delta E_2 = \frac{5E}{3} - E = \frac{2E}{3}$.The energy of the emitted photon is $\frac{hc}{\lambda'} = \frac{2E}{3}$ ... (ii)Dividing equation $(i)$ by equation (ii):$\frac{hc/\lambda}{hc/\lambda'} = \frac{2E}{2E/3}$$\frac{\lambda'}{\lambda} = \frac{2E 	imes 3}{2E} = 3$$\lambda' = 3\lambda$

The figure shows the energy levels of a certain atom. When the electron de-excites from $3E$ to $E$,an electromagnetic wave of wavelength $\lambda$ is emitted. What is the wavelength of the electromagnetic wave emitted when the electron de-excites from $\frac{5E}{3}$ to $E$?

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