A stationary He+ ion emitted a photon corresponding
to a first line of the Lyman series. The photon liberated
a photoelectron from a stationary H atom in ground
state. What is the velocity of photoelectron.
Answers
Explanation:
The first line Lyman series is between
- n=1 and n=2
. We know that
He
+
is isoelectronic with
H
atom, and thus, we can utilize the Rydberg equation for hydrogen-like atoms:
Δ
E
=
−
Z
2
⋅
R
H
(
1
n
2
f
−
1
n
2
i
)
where:
Δ
E
is the energy of the relaxed electron and thus of the emitted photon.
Z
is the atomic number.
R
H
is the Hydrogen ionization energy in the appropriate units (say,
13.61 eV
).
n
k
is the principal quantum number for the
k
th electron state.
The energy that the emitted photon has due to the electronic relaxation is given by:
Δ
E
=
−
(
2
2
)
⋅
(
13.61 eV
)
(
1
2
2
−
1
1
2
)
=
40.83 eV
and would be pertaining to the
2
p
→
1
s
−−−−−−− relaxation (due to the selection rules requiring the total orbital angular momentum to change as
Δ
L
=
±
1
).
The stationary hydrogen atom in its ground state has an ionization energy of
13.61 eV
, so its photoelectron would have a kinetic energy of:
40.83 eV
−
13.61 eV
=
27.22 eV
or
4.361
×
10
−
18
J
, using the conversion factor
1.602
×
10
−
19
J
1 eV
.
(Why did I convert to
J
?)
As all electrons have mass, they also have velocities pertaining to the usual kinetic energy equation:
K
=
1
2
m
e
v
2
e
And thus, the forward velocity is given by:
v
e
=
√
2
K
m
e
=
⎷
2
⋅
4.361
×
10
−
18
kg
⋅
m
2
/s
2
9.109
×
10
−
31
kg
=
3.094
×
10
6
m/s