Question #277268

Find the kinetic, potential, and total energies of the hydrogen atom in the first excited level, and find the wavelength of the photon emitted in the transition from the first excited level to the ground level.


1
Expert's answer
2021-12-09T09:13:26-0500

Part(a)

Total energy of first excited state

E=13.622=3.4eVE=-\frac{13.6}{2^2}=-3.4eV

For H atom Z=1

Kinetic energy H atom

K=(3.4)=3.4eVK=-(-3.4)=3.4eV

Potential energy

E=K+UU=EKE=K+U \\U=E-K

U=3.43.4=6.8eVU=-3.4-3.4=-6.8eV

Part(b)

Ground state energy

E=13.6n2E=\frac{-13.6}{n^2} eV

n=1 ground state

E=13.612=13.6eVE=\frac{-13.6}{1^2}=-13.6eV

First excited state

E=3.4eVE=-3.4eV

E=E1E2∆E=E_1-E_2


E=3.4(13.6)=10.2eV∆E=-3.4-(-13.6) =10.2eV

E=hcλ∆E=\frac{hc}{\lambda}

λ=hcE\lambda=\frac{hc}{∆E}


λ=6.625×1034×3×10810.2×1.6×1019=121.78×109=121.78nm\lambda=\frac{6.625\times10^{-34}\times3\times10^8}{10.2\times1.6\times10^{-19}}=121.78\times10^{-9}=121.78nm


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS