A photon is emitted in transition from n = 4 | vedclass

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(A) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \fr

Vedclass · Accepted Answer

$94.1$

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(A) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) \, eV$.For the transition from $n = 4$ to $n = 1$,$\Delta E = 13.6 \left( 1 - \frac{1}{16} ight) = 13.6 	imes \frac{15}{16} = 12.75 \, eV$.The energy of the photon is also given by $E = \frac{hc}{\lambda}$.Given $h = 4 	imes 10^{-15} \, eV \cdot s$ and $c = 3 	imes 10^8 \, m/s$,we have $hc = 12 	imes 10^{-7} \, eV \cdot m = 1200 \, eV \cdot nm$.Thus,$\lambda = \frac{hc}{\Delta E} = \frac{1200 \, eV \cdot nm}{12.75 \, eV} \approx 94.11 \, nm$.Therefore,the wavelength is approximately $94.1 \, nm$.

$A$ photon is emitted in transition from $n = 4$ to $n = 1$ level in a hydrogen atom. The corresponding wavelength for this transition is $......... \, nm$ (given,$h = 4 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$):

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