00 Verizon Homework: Hwk14: Classical Atom, Quan.... flipitphysics.com 12:02 PM

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00 Verizon Homework: Hwk14: Classical Atom, Quan.... flipitphysics.com 12:02 PM 1 5596.1 , Homework: Hwk14: Classical Atom, Quantum Atom And X Rays An electron in a hydrogen atom makes a transition from the n 5 energy stage to the ground state (n-1). What is the frequency of the emitted photon? Qu Now the electron is excited from its ground state to its first excited state. What is the frequency of the absorbed photon? While in this first excited state, the electron receives 2.55 eV of energy. What is the quantum number n that describes the energy level that the electron ends up at? (Enter an integer!) Copyright o 2016 Freeman Worth Publishers a division of Macmillian Learming About I Tech Support 1 Find Your Local Sales Rep Terms Of Use Privacy Palicy

Explanation / Answer

1)

energy in nth state

E = -13.6 / n^2
E(n=1) = -13.6 eV

E(n=5) = -13.6 / 25 = -0.544 eV

change in energy = 13.6 - 0.544 = 13.056 eV = emitted photon frequency = hf

f = 13.056 / h = 13.056 / 4.135 667 662 x 10-15 eV s = 3.16*10^(15) Hz

2)

now electron exited to first exited state = n =2

E(n=2) = -13.6 / 4 = -3.4 eV

so energy difference = E(n=2) - E(n=1) = 13.6 - 3.4 = 10.2 eV

frequency = f =10.2 / h = 10.2 / 4.135 667 662 x 10-15 eV s

f = 2.47*10^(15) Hz

3)

now from n = 2 it recieve energy and furthure exite and the energy difference between these new states = 2.55eV

E(n=3) = -13.6 / 9 = -1.511 eV

E(n=4) = -13.6 / 16 = -0.85 eV

E(n=5) = -13.6 / 25 = -0.544 eV

so if transition between n=2 and n=3

energy requires = E(n=2) - E(n=3) = 3.4 - 1.511 = 1.889 eV(possible)

so if transition between n=2 and n=4

energy requires = E(n=2) - E(n=4) = 3.4 - 0.85 = 2.55 eV(possible)

so transition takes palace between n=2 to n=4

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