Search any question & find its solution
Question:
Answered & Verified by Expert
$15 \mathrm{eV}$ is given to $e^{-}$in $4^{\text {th }}$ orbit, then find its final energy when it comes out of $\mathrm{H}$-atom.
Options:
Solution:
2510 Upvotes
Verified Answer
The correct answer is:
$14.15 \mathrm{eV}$
Energy of $4^{\text {th }}$ orbit of $\mathrm{H}$-atom
$=-13.6 \times \frac{1}{16}=-0.85 \mathrm{eV}$
So energy released $=$ total energy - ionization energy of $4^{\text {th }}$ orbit $=15-0.85=14.15 \mathrm{eV}$
$=-13.6 \times \frac{1}{16}=-0.85 \mathrm{eV}$
So energy released $=$ total energy - ionization energy of $4^{\text {th }}$ orbit $=15-0.85=14.15 \mathrm{eV}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.