Answer to Question #277268 in Atomic and Nuclear Physics for elie

Question

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.

Answer 1

Part(a)

Total energy of first excited state

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

For H atom Z=1

Kinetic energy H atom

$K=-(-3.4)=3.4eV$

Potential energy

$E=K+U \\U=E-K$

$U=-3.4-3.4=-6.8eV$

Part(b)

Ground state energy

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

n=1 ground state

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

First excited state

$E=-3.4eV$

$∆E=E_1-E_2$

$$∆E=-3.4-(-13.6) =10.2eV$$

$∆E=\frac{hc}{\lambda}$

$\lambda=\frac{hc}{∆E}$

$$\lambda=\frac{6.625\times10^{-34}\times3\times10^8}{10.2\times1.6\times10^{-19}}=121.78\times10^{-9}=121.78nm$$