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Physics
UTME / JAMB 1993
Modern Physics and Nuclear Physics
Solution
An element X of an atomic number 88 and mass number 226 decays to form an element Z by emitting two beta particles and an alpha particle. Z is represented by
A
\(^{222}_{82}Z\)
B
\(^{222}_{88}Z\)
C
\(^{226}_{86}Z\)
D
\(^{226}_{80}Z\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(^{226}_{88}X\) \(\to\) \((^{0}_{-1}e)\) + \(^{4}_{2}\alpha\) + \(^{b}_{a}X\)
a + 2 + (-2) = 88 a \(\Rightarrow\) a = 88
b + 4 + 0 = 266 b \(\Rightarrow\) b = 222
Source: Myschool Past Questions Archive (Record #28027)
Physics
WAEC / WASSCE 1992
Modern Physics and Nuclear Physics
Solution
An element whose half-life is 3 years has N atoms. How many atoms would have decayed after 9 years
A
1 N atoms8
B
1 N atoms3
C
2 N atoms3
D
5 N atoms6
E
7 N atoms8
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
\(\frac{N}{3}\) = 3years
\(\frac{N}{4}\) = 6years
\(\frac{N}{8}\) = 9years
N - \(\frac{N}{8}\) = \(\frac{7N}{8}\)
Source: Myschool Past Questions Archive (Record #24297)
Physics
WAEC / WASSCE 1992
Modern Physics and Nuclear Physics
When the nucleus of a uranium atom is split into two fragments of nearly equal mass the sum of the masses of the fragments is less than the mass of the original nucleus. This difference is a measure of the
A
experimental error in calculating the separate masses
B
change of momentum of each fragment
C
potential energy lost
D
nuclear energy released
E
kinetic energy lost
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #24298)
Physics
UTME / JAMB 1992
Modern Physics and Nuclear Physics
Solution
After three half lifes, the fraction of a radioactive material that has decayed is
A
\(\frac{1}{8}\)
B
\(\frac{1}{3}\)
C
\(\frac{2}{3}\)
D
\(\frac{7}{8}\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
I = \(\frac{1}{2}\) \(\Rightarrow\) \(\frac{1}{4}\) \(\Rightarrow\) \(\frac{1}{8}\)
Fraction decayed = 1 - \(\frac{1}{8}\)
= \(\frac{7}{8}\)
Source: Myschool Past Questions Archive (Record #28113)
Physics
UTME / JAMB 1992
Modern Physics and Nuclear Physics
Solution
\(^{27}_{13}Al\) + \(^{1}_{0}n\) \(\Rightarrow\) \(^{24}_{11}Na\) + X
A
a proton
B
a gamma ray
C
a helium particle
D
an alpha particle
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\(^{27}_{13}Al\) + \(^{1}_{0}n\) \(\Rightarrow\) \(^{24}_{11}Na\) + \(^{b}_{a}X\)
a + 11 = 0 + 13 \(\Rightarrow\) a = 2
b + 24 = 1 + 27 \(\Rightarrow\) b = 4
\(^{b}_{a}X\) = \(^{4}_{2}\alpha\)
Source: Myschool Past Questions Archive (Record #28114)
Physics
UTME / JAMB 1991
Modern Physics and Nuclear Physics
Solution
4g of a radioactive material of half-life 10 days is spilled on a laboratory floor. How long would it take to disintegrate 3.5g of the material?
A
1\(\frac{1}{4}\) days
B
8\(\frac{1}{4}\) days
C
30 days
D
80 days
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
half-life = 10 days
after 10 days, the 4g of element = 2g remaining with 2g disintegrated
after 10 days, 2g of element = 1g left with 1g disintegrated
hence, after 20 days, 3g of the element has been disintegrated with 1g left.
after 10 days, the 1g of element = 0.5g left with 0.5g disintegrated
after 30 days, we have 0.5g of element left with 3.5g disintegrated.
Source: Myschool Past Questions Archive (Record #28207)
Physics
UTME / JAMB 1991
Modern Physics and Nuclear Physics
Which of the following are True for \(\alpha\)-decay. i. mass number, decrease by four. ii. Atomic number decreases by two. iii. mass number does not change.
A
i and ii only
B
ii and iii only
C
i and iii only
D
i, ii, and iii only
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #28209)
Physics
WAEC / WASSCE 1991
Modern Physics and Nuclear Physics
Solution
The half-life of a radioactive substance is 2 seconds. Calculate the decay constant
A
0.035s-1
B
0.151s-1
C
0.347s-1
D
0.576s-1
E
1.386s-1
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(\lambda = \frac{0.693}{T_{\frac{1}{2}}} = \frac{0.693}{2}\)
= 0.347s-1
Source: Myschool Past Questions Archive (Record #33453)
Physics
WAEC / WASSCE 1990
Modern Physics and Nuclear Physics
Solution
A radioactive substance has a half-life of 20 hours. What fraction of the original radioactive nuclei will remain after 80 hours?
A
1/32
B
1/16
C
1/8
D
1/4
E
1/2
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
1 → 1/2 → 1/4 → 1/8 → 1/16
Source: Myschool Past Questions Archive (Record #24168)
Physics
WAEC / WASSCE 1990
Modern Physics and Nuclear Physics
Solution
Which of the following is not crystal?
A
Sodium chloride
B
Lead
C
Copper (II) tetraoxosulphate (VI)
D
Glass
E
Germanium
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Glass is not a crystal because it lacks the long-range ordered atomic structure characteristics of crystalline materials. instead, it is an amorphous solid.
Source: Myschool Past Questions Archive (Record #24172)
Physics
UTME / JAMB 1990
Modern Physics and Nuclear Physics
Solution
The number of neutrons contained in the nucleus \(^{238}_{92}U\) is
A
92
B
146
C
238
D
330
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
238 - 92 = 146
Source: Myschool Past Questions Archive (Record #28298)
Physics
UTME / JAMB 1989
Modern Physics and Nuclear Physics
Solution
Eight alpha decays and six beta decays are necessary before an atom of 2 \(^{238}_{92}U\) achieves stability. The final product in the chain has an atomic number of
A
70
B
78
C
82
D
90
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
U \(\to\) 8(\(^{4}_{2}He\)) + 6(\(^{0}_{-1}e\)) + \(^{b}_{a}X\)
92 = (8 X 2) + (6 X -1) + a
a = 82
Source: Myschool Past Questions Archive (Record #28363)
Physics
WAEC / WASSCE 1988
Modern Physics and Nuclear Physics
Solution
Two radioactive elements A and B have half-lives of 100 and 50 years respectively. Samples of A and B initially contain equal amounts/number of atoms. What is the ratio of the number of the remaining atoms of A to that of B after 200 years?
A
4 : 1
B
2 : 1
C
1 : 1
D
1 : 2
E
1 : 4
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Given that a mass N of elements A and B exists, then:
For element A, after 100 years, we have A of mass N/2 remaining
After 200 years, we have A of mass \(\frac{1}{2}\times\frac{N}{2}\) = \(\frac{N}{4}\)
For element B, after 50 years, we have B of mass N/2 remaining
After 100 years, we have B of mass \(\frac{1}{2}\times\frac{N}{2}\) = \(\frac{N}{4}\)
After 150 years, we have B of mass \(\frac{1}{2}\times\frac{N}{4}\) = \(\frac{N}{8}\)
After 200 years,we have B of mass \(\frac{1}{2}\times\frac{N}{8}\) = \(\frac{N}{16}\)
Ratio: \((\frac{N}{4} : \frac{N}{16})\)
=\(4 : 16\) = \(1 : 4\)
Source: Myschool Past Questions Archive (Record #23851)
Physics
UTME / JAMB 1987
Modern Physics and Nuclear Physics
Solution
A radioactive sample initially contains N atoms. After three half-lives the number of atoms that have disintegrated is
A
\(\frac{N}{8}\)
B
\(\frac{3N}{8}\)
C
\(\frac{5N}{8}\)
D
\(\frac{7N}{8}\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
N \(\to\) \(\frac{N}{2}\) \(\to\) \(\frac{N}{4}\) \(\to\) \(\frac{N}{8}\)
Number of atoms left = 1-1/8=\(\frac{7N}{8}\)
Source: Myschool Past Questions Archive (Record #28482)
Physics
UTME / JAMB 1985
Modern Physics and Nuclear Physics
Solution
A thin aluminum plate has a surface area of 1.500m at 20ºC. What will be its surface area when it is cooled to -20ºC? (Take the linear expansivity of aluminum to be 2.5 x 10-5k-1).
A
1.503m2
B
1.500m2
C
1.498m2
D
1.497m2
E
1.490m2
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\(\beta\) = 2\alpha
2 X2.5 x 10-5 = \(\frac{A_2 - 1.5}{1.5 \times (-20 -20)}\)
\(A_2 - 1.500\) = - 0.003m
\(A_2\) = - 0.003 + 1.500 = 1.497\(m^2\)
Source: Myschool Past Questions Archive (Record #28582)