an excited hydrogen like species when bombarded by photons of wavelength 54 nmgets excited to another excited state. in returning back to ground state a total of six different type of photons were emitted.out of 6 photons three have wavelength less than 54 nm. determine
a.initial and final excited state
b.ionization energy
c.atomic number

Dear Student,

The computation are as expressed below,

λ=54 nmE=hcλ=6.626×10-34 Js×3×108 ms-154×10-9 m=3.68×10-18 Jinitial state n=1, excited state n=61 eV = 1.602×10-19 J so we get E = 22.97 eVinitial state E=-22.97 eVn2=-22.97 eV12=-22.97 eVexcited state E=-22.97 eV62=0.638 eVIonization energy = 3.68×10-18 ×6.022×10231000=-2216.096 kJmolAtomic number (Z)=1

Regards.

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