deformere692qr

2022-05-15

Calculate the energy of the first 4 photons emitted in the balmer series of the hydrogen spectrum. The Balmer series corresponds to nf=2 and the first 4 photons are the lowest energy transitions from higher energy levels. These 4 photons all hvae a wavelength in the visible spectrum.

Raelynn Parker

To find : The energy of first 4 photons emitted in balmer series of hydrogen spectrum
Using the formula for energy of ${n}^{th}$ orbit of hydrogen :
${E}_{n}=-\frac{13.6}{{n}^{2}}eV\phantom{\rule{0ex}{0ex}}{E}_{2}=-\frac{13.6}{4}eV\phantom{\rule{0ex}{0ex}}=-3.4eV\phantom{\rule{0ex}{0ex}}{E}_{3}=\frac{-13.6}{9}eV\phantom{\rule{0ex}{0ex}}=-1.51eV\phantom{\rule{0ex}{0ex}}{E}_{4}=\frac{-13.6}{16}eV\phantom{\rule{0ex}{0ex}}=-0.85eV\phantom{\rule{0ex}{0ex}}{E}_{5}=\frac{-13.6}{25}eV\phantom{\rule{0ex}{0ex}}=-0.544eV\phantom{\rule{0ex}{0ex}}{E}_{6}=\frac{-13.6}{36}eV\phantom{\rule{0ex}{0ex}}=-0.377eV$
A) Energy of first photon of balmer series
$={E}_{3}-{E}_{2}\phantom{\rule{0ex}{0ex}}=-1.51-\left(-3.4\right) eV\phantom{\rule{0ex}{0ex}}=1.89eV$
B) Energy of second photon of balmer series $={E}_{4}-{E}_{2}\phantom{\rule{0ex}{0ex}}=-0.85-\left(-3.4\right) eV\phantom{\rule{0ex}{0ex}}=2.55eV$
C)Energy of third photon of balmer series $={E}_{5}-{E}_{2}\phantom{\rule{0ex}{0ex}}=-0.544-\left(-3.4\right) eV\phantom{\rule{0ex}{0ex}}=2.856eV$
D) Energy of fourth photon of balmer series $={E}_{6}-{E}_{2}\phantom{\rule{0ex}{0ex}}=-0.377-\left(-3.4\right) eV\phantom{\rule{0ex}{0ex}}=3.02eV$

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