Determine the de-broglie wavelength associated with he atom at 500k
Answers
Answer:
I got
0.0794 nm
, or
79.4 pm
, over twice its atomic radius.
Well, assuming that the
He
atoms are all moving in the same direction, and follow a Maxwell-Boltzmann distribution:
f
(
u
)
=
4
π
(
m
2
π
k
B
T
)
3
/
2
u
2
e
−
m
u
2
/
2
k
B
T
they thus have the average speed
u
a
v
g
=
∫
∞
0
u
f
(
u
)
d
u
=
[
.
.
.
]
=
√
8
k
B
T
π
m
,
where:
k
B
=
1.38065
×
10
−
23
is the Boltzmann constant in
J/K
, or
kg
⋅
m
2
/s
2
⋅
K
.
T
is the temperature in
K
.
m
is the per-particle mass in
kg
, i.e.
M
/
N
A
=
m
, where
M
is the molar mass in
kg/mol
and
N
A
=
6.0221413
×
10
23
mol
−
1
.
We then use this average speed as the velocity
v
in the de Broglie relation
λ
=
h
m
v
,
where
h
=
6.626
×
10
−
34
J
⋅
s
, or
kg
⋅
m
2
/s
, is Planck's constant and
m
is as defined before,
since the particles are assumed to all be moving in the same direction so that the velocity is the speed.
Therefore, the wavelength is:
λ
=
h
m
×
√
π
m
8
k
B
T
=
√
π
h
2
8
m
k
B
T
=
⎷
π
(
6.626
×
10
−
34
kg
⋅
m
2
/
s
)
2
8
(
0.0040026
kg
/
mol
×
1 mol
6.0221413
×
10
23
)
(
1.38065
×
10
−
23
kg
⋅
m
2
/
s
2
⋅
K
)
(
298.15
K
)
=
7.938
×
10
−
11
m
=
0.0794 nm
,
which is in the gamma-ray region of the EM spectrum