When a hydrogen atom is excited by radiation | vedclass

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(B) The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{12400 \, 	ext{eV} \cdot \mathring{A}}{975 \, \mathring{A}} \approx 12.718 \,

Vedclass · Accepted Answer

$18787 \, \mathring{A}, 975 \, \mathring{A}$

Vedclass · Answer

(B) The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{12400 \, 	ext{eV} \cdot \mathring{A}}{975 \, \mathring{A}} \approx 12.718 \, 	ext{eV}$.This energy corresponds to the transition from the ground state $(n=1)$ to the $n=4$ state,as $E_4 - E_1 = -0.85 - (-13.6) = 12.75 \, 	ext{eV}$.When the atom de-excites from $n=4$,the possible transitions are $4 	o 3, 4 	o 2, 4 	o 1, 3 	o 2, 3 	o 1, 2 	o 1$.The maximum wavelength corresponds to the minimum energy transition,which is $4 	o 3$.$E_{4 	o 3} = E_4 - E_3 = -0.85 - (-1.51) = 0.66 \, 	ext{eV}$.$\lambda_{max} = \frac{12400}{0.66} \approx 18787.8 \, \mathring{A}$.The minimum wavelength corresponds to the maximum energy transition,which is $4 	o 1$.$E_{4 	o 1} = 12.75 \, 	ext{eV}$.$\lambda_{min} = \frac{12400}{12.75} \approx 972.5 \, \mathring{A} \approx 975 \, \mathring{A}$ (as per the excitation energy).Thus,the values are $18787 \, \mathring{A}$ and $975 \, \mathring{A}$.

When a hydrogen atom is excited by radiation of wavelength $975 \, \mathring{A}$,the maximum and minimum wavelengths of the emitted radiation are respectively:

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