In hydrogen-like atom (atomic number A is in a higher excited state of quantum number n. This excited atom can make a transition to first excited state by emitting photons of energies 10.20 eV and 17.00 eV respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine values of n (ionization energy of hydrogen atom = 13.6 eV).
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
answer is 6.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Given that En−E2=10.2+17.0=27.2eV and En−E3=4.25+5.95=10.2eV⇒E3−E2=27.2−10.2=17eV⇒1.89Z2=17eV⇒Z=3 Now we use 27.2=13.6Z214−1n2⇒n=6
Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!
In hydrogen-like atom (atomic number A is in a higher excited state of quantum number n. This excited atom can make a transition to first excited state by emitting photons of energies 10.20 eV and 17.00 eV respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine values of n (ionization energy of hydrogen atom = 13.6 eV).