A Hydrogen atom has one proton in the nucleus and one electron in the shell. In
ID: 1384097 • Letter: A
Question
A Hydrogen atom has one proton in the nucleus and one electron in the shell. In a classic model of the atom, in a certain state, this electron is in a circular orbit around the nucleus with an velocity of 545364.928909953 m/s.
What is the radius of the orbit?
What is the angular momentum, L, of the electron at this radius?
What is the quantum value, n, of the electron at this radius?
What is the total energy of the electron at this radius?
What is the total energy of the electron at this radius in electron volts?
If someone could explain this to me, I'd really appreciate it!
Explanation / Answer
we know Angular momentum of elelctron is conserved.
L = n*h/(2*pi)
n = L*2*pi/h
= 5.275*10^-34*2*pi/(6.625*10^-34)
= 5 (principe quantum number)
so,
In hydrogen atom,
radius of orbit, rn = ro*n^2 (here ro = 0.0529 nm is the Bhor radius(first orbit radius)
r5 = 0.0529*10^-9*5^2
= 1.325*10^-9 m
Use, L = m*v*r
==> v = L/(m*r)
= 5.275*10^-34/(9.1*10^-31*1.33*10^-9)
= 436000 m/s
KE = 0.5*m*v^2
= 0.5*9.1*10^-31*(436000)^2
= 8.65*10^-20 J
KE(in eV) = 8.65*10^-20/(1.6*10^-19)
= 0.54 eV
qunatum number n = 5
Total Energy of the elctron at this radius = -13.6/5^2 = -0.544 eV
Total Energy in jouls = -0.544*1.6*10^-19 = -8.7*10^-20 J
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