A hydrogen atom is in its second excitedstate ( n = 3). Using the Bohrtheory of
ID: 1679528 • Letter: A
Question
A hydrogen atom is in its second excitedstate (n = 3). Using the Bohrtheory of the atom, calculate the following. (a) the radius of the orbit1 nm
(b) the linear momentum of the electron
2 kg·m/s
(c) the angular momentum of the electron
3 J·s
(d) the kinetic energy
4 eV
(e) the potential energy
5 eV
(f) the total energy
6 eV thank you in advance. (a) the radius of the orbit
1 nm
(b) the linear momentum of the electron
2 kg·m/s
(c) the angular momentum of the electron
3 J·s
(d) the kinetic energy
4 eV
(e) the potential energy
5 eV
(f) the total energy
6 eV thank you in advance.
Explanation / Answer
in the given problem n = 3
the conversion is
1 eV = 1.602 x 10-19 J
(a)
rn = n2 ao
where ao = 0.0529 nm
rn = ............ nm
(b)
The Linear momentum of the electron is given by
pn = mvn
= m ( 2ke2 /nh )
For n =3, we have p3= m ( 2ke2 /3h )
=2(9.11*10-31kg)(9*109Nm2/C)(1.60*10-19C)/ [3(6.63*10-34J.s)]
=6.62*10-25kgm/s
c) The angular moemntum of the electron
Ln = nh / 2
therefore, L3= 3h / 2
substitute and calculate the value ofL3
d) Therefore the KE for n = 3 is
KE3 = E3 - PE3
e) The PE of the electron in the n=3 orbit is
PE3 = - kZe2 / R3
where the radius R3 =9R1
= 9*(0.529*10-10m)
and the atomic number Z =1 for hydrogen
f) The Total energy for the 3rd orbit is isgiven by
E3 = - 13.6eV / 9 = -1.51eV
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