PGE OUR VALUE 2021
PHY3243: Atomic, Particles and Nuclear Physics
Assignment 1
Date: 27/10/2020
Due date: 2/11/2020
1. An X-ray tube operates at 15 × 104 𝑉 and 10𝑚𝐴. (a) If only 1% of the electric power supplied is
converted into X-ray, at what rate is the target being heated in Calories per second? (b) If the
target weights 300gr and has a specific heat of 0.035 𝑐𝑎𝑙/𝑔𝑜 𝑐, at what average would its
temperature rise if there were no thermal loses? (c) What must be the physical properties of a
practical target material? What would be some suitable elements?
Answer:
a) Heat rate (heat energy per second) (Joule effect):
𝑃 = (𝑉𝐼)(0.99) = 15 × 104 × 10 × 10−3 × 0.99 = 1485𝐽/𝑠
or
1485𝐽/𝑠
= 354.754 𝐶𝑎𝑙/𝑠
4.186𝐽/𝐶𝑎𝑙
(1 𝐶𝑎𝑙 = 4.186𝐽) Answer
b) ∆𝑄 = 𝑚𝑐∆𝑇 or the rate of change of heat is ∆𝑄/∆𝑡 = 𝑚𝑐∆𝑇/∆𝑡 or 𝑃 = 𝑚𝑐∆𝑇/∆𝑡
Where ∆𝑇/∆𝑡 is rate of change of temperature.
∆𝑇
𝑃
354.754𝐶𝑎𝑙/𝑠
Hence, the rise in temperature per second, ∆𝑡 = 𝑚𝑐 = 300𝑔×0.035𝐶𝑎𝑙/𝑔𝑜 𝐶 = 33.79𝑜 𝐶/𝑠
c) The melting point or specific heat capacity of the target should be high enough.
-The suitable elements are: Tungsten, Cobalt, Copper, etc.
74
Ex: Tungsten 184
𝑊, ha a melting point of 3406.85𝑜 𝐶 and
3 marks
Specific heat capacity of 0.132𝐽/𝑔𝑜 𝐶
2. X-ray from a certain cobalt target tube are composed of the strong 𝐾 series of Cobalt and weak
𝐾 lines due to impurities. The wavelengths of 𝐾𝛼 are 1.785𝐴𝑜 for Cobalt and 2.285𝐴𝑜 and
1.537𝐴𝑜 for impurities. (a) What elements are they? [For the 𝐾-series, a=1 in Moseley’s law]
Answer:
1
Mosley’s law says that: 𝜆 = 𝐶(𝑍 − 𝑎)2
for 𝐾-serries, 𝑎 = 1
𝜆 = 1.785𝐴𝑜
𝑍(Cobalt)= 27
1
The constant 𝐶 = 1.785×10−10 ×(26)2
Or 𝐶 = 0.8287 × 107 𝑚−1
1
a) (𝑍 − 1)2 = 𝐶𝜆 or 𝑍 = 1 +
1
√𝐶𝜆
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For 𝜆 = 2.285𝐴𝑜 , 𝑍 = 1 + 22.98
or 𝑍 = 24 which is 𝐶𝑟-Cromium
b) For 𝜆 = 1.537𝐴𝑜 𝑍 = 1 + 28.02 or 𝑍 = 29 (𝐶𝑢-Copper)
2 marks
3. Calculate the energies required to create a 𝐾-vacancy and 𝐿-vacancy in a copper atom (𝑍 = 29).
Obtain the wavelength of the 𝐾𝛼 -line
Answer:
(a) The energy needed to create a vacancy with 𝐾-shell (𝑛 = 1) is:
𝑍 2 × 13.64 𝑒𝑉 = 292 × 13.64 𝑒𝑉 = 11.5 𝐾𝑒𝑉
The energy required to create a vacancy in the 𝐿-shell (𝑛 = 2), assuming that 𝑍𝑛 = 27 (we
remove 2es because they are in K-shell and the remaining charge screened by these 2 es is
Z= 27) is:
𝑍𝑛2 × 13.64 272 × 13.64
=
= 2.5 𝐾𝑒𝑉
𝑛2
4
Using the Moseley’s law:
1
3
= 4 𝑅𝑀 (𝑍 − 1)2
𝜆
4
1
3
1
or 𝜆 = 3 𝑅 (𝑍−1)2 = 4 1.097×107 ×(28)2 or 𝜆 = 1.55 × 10−10 𝑚 = 1.55𝐴𝑜
𝑀
2 marks
4. The table below gives the 𝐾𝛼 -lines for various elements. (a) Determine the atomic number of
each element from data; (b) What elements are they? [Use 𝑅𝑀 = 1.09737 × 107 𝑚−1 ]
𝜆 of 𝐾𝛼 line in 𝐴𝑜
Element
A
B
C
D
1.79
1.66
1.54
1.43
Answer:
Mosley law for 𝐾𝛼 -lines:
1
3
4 1
= 4 𝑅𝑀 (𝑍 − 1)2 or 𝑍 = 1 + √3 𝜆𝑅
𝜆
𝑀
𝑅𝑀 = 1.09737 × 107 𝑚−1
A: 𝜆 = 1.97𝐴𝑜 = 1.97 × 10−10 𝐴𝑜 𝑚
𝑍 = 1 + 26.07 ≅ 27, 𝐶𝑜-Cobalt
B: (𝜆 = 1.97𝐴𝑜 ), 𝑍 = 1 + 27.05 ≅ 28, 𝑁𝑖-Nickel
C: (𝜆 = 1.54𝐴𝑜 ), 𝑍 = 1 + 28.09 ≅ 29, 𝐶𝑢-Copper
D: (𝜆 = 1.43𝐴𝑜 ), 𝑍 = 1 + 29.15 ≅ 30, 𝑍𝑛-Zinc
4 marks
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5. (i) From the basics principles of Bohr atomic model, derive an expression for the energy
of hydrogen atom. Use standards values in SI units to compute that energy and show
that it is quantized. (ii)Compute the energy of hydrogen atom in its first excited state.
answer
Energy of hydrogen atom by Bohr’s atomic model
An electron evolving on its orbit has a centripetal force
V2
(a)
Fc = m
r
Ze2
And Coulomb force Fc =
(b)
4πε0 r 2
V2
Ze2
(a) = (b) <=> 𝑚
=
r
4πε0 r 2
nℏ
mvr = nℏ => 𝑉 =
mr
2 2
4πε0 n ℏ
r=
(2)
mZe2
nℏ
( 1)
Ze2
V = mr = 4πε nℏ
0
The total mechanical energy is
E = KE + V
1
(𝟑)
E = mv 2 + V
where V is the pontial energy
2
Rearrangement (𝟏)(𝟐), 𝐚𝐧𝐝 (𝟑)
mZ z e4
En = −
is the expression, n = 1,2,3, …
(4πε0 )2 2ℏ2 n2
By substituting each constant,
En = −
13.6Z2
n2
in eV
This is the quantized energy
(ii) For first excited state, n=2
En = −
13.6×12
22
13.6
= − 4 = −3.4eV
3 marks
6. Electron are accelerated in a television tube through a potential difference of 9.8kV.
Find the highest frequency and minimum wavelength of the electromagnetic wave
emitted, when these strike the screen of the tube. In which region of the spectrum will
these lie?
a. Differentiate X-rays and visible light in terms of the origin of these EM radiations.
b. What voltage must be applied to an X-ray tube for it to emit X-rays with minimum
wavelength 0.5 A˚ ?
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Benjamin
Answer:
The highest frequency that corresponds to minimum wavelength is:
ℎ𝑐
𝐾 = ℎ𝜈𝑚𝑎𝑥 =
= 9.8𝐾𝑒𝑉
𝜆𝑚𝑖𝑛
9.8𝐾𝑒𝑉
9.8 × 103 𝑒𝑉
𝜈𝑚𝑎𝑥 =
=
= 2.37 × 1018 𝑠 −1
ℎ
4.135 × 10−15 𝑒𝑉𝑠
3×108 𝑚𝑠 −1
Or 𝜆𝑚𝑖𝑛 = 2.37×1018𝑠−1 = 1.265 × 10−10 𝑚 = 1.265A˚
The spectrum will lie in the X-ray region.
a) While visible light involve wavelength range of 400𝑛𝑚 to 700𝑛𝑚 and originate
from electron transition of outer atomic levels that require energy of few electron
volts, the X-rays involve spectrum of wavelength range of 1A˚ and originate from
electron transition of inner atomic levels and require energy in the range of Kilo
electron volts.
b) As 𝐾 = 𝑒𝑉 = 𝜆
ℎ𝑐
𝑚𝑖𝑛
ℎ𝑐
or 𝑉 = 𝑒𝜆
𝑚𝑖𝑛
4.135×10−15 𝑒𝑉𝑠×3×108 𝑚𝑠 −1
=
𝑒5×10−11 𝑚
= 2.4813 × 104 𝑉
Or the voltage to be applied to produce X-rays with minimum wavelength of 0.5A˚
3 marks
is 24.813𝐾𝑉
7. Find the energy in eV and wavelength of Kα X-ray line of 27Co.
Answer:
Kα X-ray line involve transition energy from E2 to E1
i)
Using Mosley law, 𝐸𝑛 = −13.6𝑒𝑉
(𝑍−1)2
𝑛2
For X-ray, 𝐾𝛼 involve the transition from 𝐸2 to 𝐸1 and
therefore, ∆𝐸 = 𝐸2 − 𝐸1 = −13.6𝑒𝑉 (
(𝑍−1)2
22
−
(𝑍−1)2
12
1.5 marks
10.2𝑒𝑉(26)2 = 6895.2𝑒𝑉 = 6.89𝐾𝑒𝑉
ii)
ℎ𝑐
For wavelength, 𝜆 = 𝐸 =
) = 10.2𝑒𝑉(𝑍 − 1)2 =
4.14×10−15 ×3×108
6895.2
= 1.8 × 10−10 𝑚 = 1.8𝐴𝑜
1.5marks
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