why the frequencies of the x-rays is directly proportional to the atomic number of the element ? (Henry Moseley)
Answers
Answer:
The frequency
ν
ν of a characteristic X-ray of an element is related to its atomic number
Z
Z by
√
ν
=
a
(
Z
−
b
)
,
ν=a(Z−b),
where
a
a and
b
b are constants called proportionality and screening (or shielding) constants. For
K
K series, the value of
a
a is
√
3
R
c
/
4
3Rc/4 and that of
b
b is 1. Here
R
R is Rydberg's constant and
c
c is speed of light (as in Bohr's model). For
L
L series, the value of
a
a is
√
5
R
c
/
36
5Rc/36 and
b
b is 7.4. The relation and values of
a
a and
b
b are experimentally determined by Henry Moseley.
Moseley's Law Formula.
Relation between the frequency of characteristic x-ray and the atomic number Z. The line intersect the Z axis at
Z
=
b
Z=b (b is 1 for K series and it is 7.4 for L series).
Solved Problems on Moseley's Law
Problem from IIT JEE 2003
Characteristic X-rays of frequency
4.2
×
10
18
4.2×1018 Hz are produced when transitions from
L
L-shell to
K
K-shell take place in a certain target material. Use Moseley's law to determine the atomic number of the target material. (Rydberg's constant
=
1.1
×
10
7
m
−
1
=1.1×107m−1.)
Solution:
The characteristic X-ray is emitted when an electron in
L
L shell makes a transition to the vacant state in
K
K shell. In Moseley's equation,
√
ν
=
a
(
Z
−
b
)
,
ν=a(Z−b),
the parameter
b
≈
1
b≈1 for this transition because electron from
L
L shell finds nuclear charge
Z
e
Ze shielded by remaining one electron in
K
K shell i.e., effective nuclear charge is
(
Z
−
1
)
e
(Z−1)e. Thus, by substituting values,
1
λ
=
ν
c
=
4.2
×
10
18
3
×
10
8
=
R
(
Z
−
1
)
2
[
1
n
2
1
−
1
n
2
2
]
=
1.1
×
10
7
(
Z
−
1
)
2
[
1
1
2
−
1
2
2
]
,
1λ=νc=4.2×10183×108=R(Z−1)2[1n12−1n22]=1.1×107(Z−1)2[112−122],
which gives,
Z
=
42
Z=42. Moseley's law played key role in arrangement of elements in the periodic table and to find many new (missing) elements.
Problem from IIT JEE 2014
If
λ
C
u
λCu is the wavelength of
K
α
Kα X-ray line of copper (atomic number 29) and
λ
M
o
λMo is the wavelength of the
K
α
Kα X-ray line of molybdenum (atomic number 42), the the ratio
λ
C
u
/
λ
M
o
λCu/λMo is close to
1.99
2.14
0.50
0.48
Solution:
The wavelength of
K
α
Kα X-ray line is related to atomic number
Z
Z by Moseley's Formula
1
λ
=
R
(
Z
−
1
)
2
[
1
1
2
−
1
2
2
]
=
3
4
R
(
Z
−
1
)
2
.
1λ=R(Z−1)2[112−122]=34R(Z−1)2.
Substitute the value of
Z
Z to get
λ
C
u
λ
M
o
=
(
Z
M
o
−
1
)
2
(
Z
C
u
−
1
)
2
=
(
41
)
2
(
28
)
2
=
2.14.
λCuλMo=(ZMo−1)2(ZCu−1)2=(41)2(28)2=2.14.
The elements with higher atomic number (molybdenum in this example) gives high energy X-rays (short wavelengths).
Problem from IIT JEE 2008
Which of the following statements is wrong in the context of X-rays generated from a X-ray tube?
Wavelength of characteristic X-rays decreases when the atomic number of the target increases.
Cut-off wavelength of the continuous X-rays depends on the atomic number of the target.
Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube.
Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube.
Solution:
The frequency
ν
ν of characteristic X-rays is related to atomic number
Z
Z by Moseley's law,
√
ν
=
a
(
Z
−
b
)
,
ν=a(Z−b),
which gives
λ
=
c
ν
=
c
a
2
(
Z
−
b
)
2
.
λ=cν=ca2(Z−b)2.
Thus, the wavelength of emitted X-rays decreases with increase in
Z
Z. The cut-off wavelength of continuous X-rays corresponds to maximum energy of electron in X-ray tube. It is given by
h
c
/
λ
=
e
V
,
hc/λ=eV,
where
V
V is the accelerating potential. The intensity of X-rays depends on the number of electrons striking the target per second, which, in turn, depends on the electrical power given to the X-ray tube as energy of each electron is
e
V
eV.
Questions on Moseley's Law
Question: If 178.5 pm is the wavelength of X-ray line of copper (atomic number 29) and 71 pm is the wavelength of the X-ray line of molybdenum (atomic number 42) then the value of a and b in Moseley's equation are
A.
a
=
7
×
10
7
H
z
1
/
2
a=7×107Hz1/2, b=1.0
B.
a
=
5
×
10
7
H
z
1
/
2
a=5×107Hz1/2, b=7.4
C.
a
=
5
×
10
7
H
z
1
/
2
a=5×107Hz1/2, b=1.4
D.
a
=
7
×
10
7
H
z
1
/
2
a=7×107Hz1/2, b=7.4
Select and click to check answer
Question: Moseley's Law for characteristic X-rays is
√
ν
=
a
(
Z
−
b
)
ν=a(Z−b). In this formula
A. both a and b are independent on the material.
B. a is independent but b depends on the material.
C. b is independent but a depends on the material.
D. both a and b depend on the material.
Select and click to check answer
Related Topic
Characteristic and Continuous X-rays | Problems | IIT JEE
References and External Links
Concepts of Physics Part 2 by HC Verma (Link to Amazon)
IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
X-ray Fluorescence and Moseley’s Law (pdf article on X-rays, Moseley's Law and Moseley's Experiments)
The Physical (in)significance of Moseley's Screening Parameters, (journal article in pdf by K Razi Naqvi)
One hundred years of Moseley’s law: An undergraduate experiment with relativistic effects (pdf)