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Physics 102 Fall 2019 Chapter 29 Homework roblem 29.16 - Enhanced - with Feedback The allowed energies of a simple atom are 0.0 V.4.0 eV, and 6.0 V. Part A What wavelength(s) appear(s) in the atom's emission spectrum? Express your answer in nanometers. If there is more than one answer, enter your answers in ascending order separated by commas 90 AERO ? Hemiskon 310,207 Submit Previous Answers Request Answer Part B What wavelength(s) appears in the stom's absorption spectrum? Express your answer in nanometers if there is more than one answer, enter your answers in ascending order separated by commas. AYDO ? plan 1621 Submit Previous Anime Ranest Answer O type here to QUVVd-31MBH

Answer 1

Solution:

The allowed energies are : 0 eV, 4 eV and 6 eV.

The emission of the photon will be for either of this transitions:

6 eV - 0 eV = 6 eV

4 eV - 0 eV = 4 eV

6 eV - 4 eV = 2 eV

So, the wavelength ($\lambda$) = h c / E = {(6.6 x 10^-34)(3 x 10⁸ )}/ (1.6 x 10^-19) n = 12.375 x 10^-7 / n

$\lambda$₁ = 2.0625 x 10^-7 m = 206.25 nm or 210 nm approx.

$\lambda$₂ = 3.09375 x 10^-7 m = 309.38 nm or 310 nm approx.

$\lambda$₃ = 6.1875 x 10^-7 m = 618.75 nm or 620 nm approx.

Part (B)Answer : 206.25 nm and 309.38 nm { or 210 nm and 310 nm approx. }