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Multiple choice
An electron (
m
=
9.109
e
−
31
m = 9.109e-31
m
=
9.109
e
−
31
kg) is accelerated through
V
=
54
V = 54
V
=
54
V (the Davisson-Germer energy), so
p
=
2
m
e
V
p = \sqrt{2 m e V}
p
=
2
m
e
V
with
e
=
1.602
e
−
19
e = 1.602e-19
e
=
1.602
e
−
19
C. Its de Broglie wavelength is
λ
=
h
/
p
\lambda = h/p
λ
=
h
/
p
with
h
=
6.626
e
−
34
h = 6.626e-34
h
=
6.626
e
−
34
J*s. What is
λ
\lambda
λ
in picometers (
1
1
1
pm
=
1
e
−
12
= 1e-12
=
1
e
−
12
m)?
🔬 try it before you answer
166.9 pm
1.669 pm
333.8 pm
83.4 pm
Check answer