1
Chem 6 Sample Exam 2 Brief Answers
1a
.
Two possible approaches here to find the energy difference and hence the
frequency.
Either find the energy of both states and subtract, or use one of the formulas
that includes the subtraction for you.
∆
E = h
ν
The energy of the n = 253 and n = 252 states is given by
E
n
= –2.18x10
–18
J(Z
2
/n
2
) where Z = 1 for the hydrogen atom
E
253
= –2.18x10
–18
J(1
2
/253
2
) = –3.406x10
–23
J
E
252
= –2.18x10
–18
J(1
2
/252
2
) = –3.433x10
–23
J
∆
E = E
final
– E
initial
= E
252
– E
253
= –2.7x10
–25
J (a negative number, so energy is emitted)
h
ν
=
∆
E so
ν
= l
∆
El/h = 2.7x10
–25
J/6.626x10
–34
Js =
4.1x10
8
s
–1
or 410 MHz
(radiofrequency)
your final numerical result may be somewhat different, depending on how you rounded
off the numbers above
Or,
ν
= 3.29x10
15
s
–1
Z
2
(1/n
f
2
– 1/n
i
2
).
Plug in Z = 1, n
f
= 252, n
i
= 253, to get
ν
= 4.1x10
8
s
–1
1b
. The Bohr radius is given by r = a
o
n
2
/Z Å where a
o
= 0.529 Å = 0.529x10
–10
m
Plug in n = 253 and Z =1 to get
r = 0.529x10
–10
m(253)
2
=
3.39x10
–6
m
x 1000mm/m = 3.39x10
–3
mm.
That seems a bit small to see unless
you've got really good vision.
But still, mighty big for an atom!
2
. E
n
= n
2
h
2
/8mL
2
Plug in n=1 and 2, plus all the constants
E
1
= (6.626x10
–34
Js)
2
(1)
2
/8(9.11x10
–31
kg)(1.34x10
–10
m)
2
= 3.35x10
–18
J
E
2
= same thing, but multiplied by 2
2
instead of 1
2
, so it's 1.34x10
–17
J
The energy difference is
∆
E= IE
2
–E
1
I = 1.01x10
–17
J
E = h
ν
= hc/
λ
so
λ
= hc/E = (6.626x10
–34
Js)(3x10
8
m/s)/ 1.01x10
–17
J
= 1.97x10
–8
m x 10
9
nm/m = ~
20 nm
.
You could also do this by using the formula for the energy difference between states of
quantum numbers n and n+1:
∆
E = (2n+1)h
2
/8mL
2
, plugging in n=1 to get the separation between E
1
and E
2
.
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2
3
. Zeropoint energy is defined as the minimum energy of the system.
So for the H atom
it's the energy of the ground state, where n =1, given by
E
n
= –2.18x10
–18
JZ
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 Spring '08
 Gribble
 Atom, Electron, Ion, Chemical bond, greater radial extent

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