In a hydrogen like sample,
electron is in 2nd excited
state, the Binding energy o
4th state of this sample is
13.6 eV then
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Answer:
Explanation:
Given data
The energy in
n
=
4
t
h
n=4th state is
E
=
13.6
e
V
E=13.6eV.
The expression for the energy is,
E
=
13.6
z
2
n
2
E=13.6z2n2
Here,
z
z is the atomic number.
Substitute the values.
13.6
e
V
=
13.6
z
2
4
2
z
2
=
4
2
z
=
4
13.6eV=13.6z242z2=42z=4
The sample for the
z
=
4
z=4 will be Berellium (Be).
The energy to set free the electron from 2nd excited state
(
n
=
3
)
(n=3) of Be is,
E
′
=
13.6
(
4
)
2
(
3
)
2
=
24.177
e
V
E′=13.6(4)2(3)2=24.177eV
Thus, a 25 eV photon can set free the electron from the second excited state of the sample.
The number of spectral lines when electron transition takes place from second excited state
(
n
=
3
)
(n=3) to ground state
(
n
=
1
)
(n=1) is,
n
=
(
3
−
1
)
(
3
−
1
+
1
)
2
=
3
n=(3−1)(3−1+1)2=3
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