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Physics
WAEC / WASSCE 2024
Structure of the Atom
Solution
What are cathode rays?
A
Stream of electrons
B
Beam of neutrons
C
Beam of protons
D
Stream of alpha particles
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Cathode rays are streams of electrons emitted from the cathode (negative electrode) in a vacuum tube. Cathode rays can be deflected by magnetic and electric fields, demonstrating their charge and nature as electrons.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72012)
Physics
WAEC / WASSCE 2024
Structure of the Atom
Solution
Light of energy 1.12 x 10\(^{-18}\) J is incident on a metal of ejected electrons with maximum energy of 8.0 x 10\(^{-19}\)J. Calculate the threshold frequency of the metal. [h = 6.626 x 10\(^{-34}\)Js]
A
4.8 x 10\(^{14}\)Hz
B
1.2 x 10\(^{15}\)Hz
C
1.7 x 10\(^{15}\)Hz
D
2.8 x 10\(^{15}\)Hz
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
E = hf + W\(_o\)
E = 1.12 x 10\(^{-18}\)J
Max. Energy = 8.0 x 10\(^{-19}\)J.
W\(_o\) = E - Max. Energy = 1.12 x 10\(^{-18}\)J - 8.0 x 10\(^{-19}\)J. = 3.2 x 10\(^{-19}\)J.
But W\(_o\) = hf\(_o\)
f\(_o\) = \(\frac{W_o}{h}\) = \(\frac{3.2 \times 10^{-19}}{6.626 \times 10^{-34}}\) = 4.83 x 10\(^{14}\)Hz
Source: WAEC / JAMB / NECO Classroom Archive (Record #72058)
Physics
WAEC / WASSCE 2024
Structure of the Atom
Solution
(a) Mention three facts about photoelectric effect
(b) An electric magnetic radiation source of power 6 x 10\(^{-3}\) W emits 1 x 10\(^{16}\) photons per second. The most energetic photo electron ejected from a metal surface is stopped by a potential difference of 2.2V. Calculate the work function of the metal. [ mass of (photon) electrons 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C]
(c) State two factors on which the activity of a radioactive sample depends
(d) Cobalt-60 source has an activity of 2.0 x 10\(^6\) Bq and a half-life of 1.8 x 10\(^8\)s. Calculate the number of radioisotope nuclei in the source.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) electrons are only emitted when the photon frequency exceeds the threshold frequency,
- the energy of each photon is proportional to its frequency,
- the emission of photoelectrons depends on the frequency of incident light on the metal,
- the intensity of light does not affect the energy of the electrons emitted/photoelectrons,
- the energy of the photoelectron depends on the frequency of incident radiation.
(b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ?
Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\)
= \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\)
= 6 x 10\(^{-19}\)J
But K.E = E + W\(_o\)
W\(_o\) = E - K.E = E - eV
W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J
(c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc
(d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s
A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity.
But \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\)
\(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\)
Recall, A = \(\lambda\)N
N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei
Source: WAEC / JAMB / NECO Classroom Archive (Record #72095)
Physics
UTME / JAMB 2024
Semi Conductor
Solution
A rectifier is a device that changes
A
D.C to D.C
B
D.C to A.C
C
A.C to A.C
D
A.C to D.C
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
A rectifier is a device that converts alternating current (AC) into direct current (DC). It enables the flow of current in one direction while blocking it in the opposite direction, thus transforming AC voltage into a form of DC voltage suitable for different applications.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70370)
Physics
UTME / JAMB 2024
Semi Conductor
Solution
The process of adding impurities to a semiconductor material to increase its conductivity is
A
Rectification
B
Emission
C
Doping
D
Stabilization
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Doping involves introducing specific impurities into a semiconductor to modify its electrical properties, enhancing its conductivity by creating either n-type or p-type semiconductors.
Rectification is the process of converting alternating current (AC) into direct current (DC) using diodes, which allow current to flow in one direction.
Emission refers to the release of charge carriers (electrons or holes) from a semiconductor, often resulting in the production of light in devices like LEDs
Stabilization involves techniques to maintain the consistent performance of semiconductor devices under varying conditions.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70422)
Physics
UTME / JAMB 2024
Semi Conductor
Solution
Which of the following materials has a very large energy gap band?
A
Conductor
B
Insulator
C
Semiconductor
D
Intrinsic semiconductor
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Conductor: Has a very small or negligible band gap, allowing electrons to move freely.
Insulator: Has a large energy gap (typically greater than 3 eV), which prevents electrons from jumping from the valence band to the conduction band, resulting in very low electrical conductivity.
Semiconductor: A moderate energy gap (about 1 eV) allows some conductivity under certain conditions (like temperature or doping).
Intrinsic Semiconductor: A type of semiconductor with a pure form, that also has a moderate energy gap similar to other semiconductors.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70478)
Physics
WAEC / WASSCE 2024
Structure of the Nucleus
Solution
All of the following uses are practical applications of x-ray except
A
the treatment of tumors
B
determining the depth of a mine
C
dental operation
D
detecting fractures in bones
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
X-rays are typically used for medical imaging and other applications related to examining structures, but they are not used to determine the depth of a mine
Source: WAEC / JAMB / NECO Classroom Archive (Record #72046)
Physics
WAEC / WASSCE 2024
Structure of the Nucleus
Solution
The main difference between X-rays and gamma rays is in their
A
ionizing ability.
B
absorbing rate.
C
mode of propagation.
D
mode of production.
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
X-rays are usually produced by electronic transitions within atoms, whereas gamma rays are released from an atom's nucleus during radioactive decay.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72069)
Physics
WAEC / WASSCE 2024
Structure of the Nucleus
Solution
The correct sequence of ionization power of alpha particle, gamma ray, and beta particle respectively is
A
low, medium, high.
B
high, medium, low.
C
medium, low, high.
D
high, low, medium.
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The correct sequence of ionization power from highest to lowest is: Alpha particle > Beta particle > Gamma ray.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72072)
Physics
WAEC / WASSCE 2024
Wave Particle Paradox
Solution
An electron of mass, m, and charge, e moves through the electric field of potential difference V\(_o\) with a speed, v. Show that de Broglie wavelength associated with the electron is given as \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\), where h is the Plank's constant.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
To show that \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
The de broglie wavelength \(\lambda\), is given by
\(\lambda\) = \(\frac{\text{h}}{\text{p}}\)
where p = mv
\(\lambda\) = \(\frac{\text{h}}{\text{mv}}\)
When an electron is accelerated through a potential difference V\(_0\), it gains kinetic energy given K.E.= eV\(_0\)
The kinetic energy can also be expressed in terms of the mass m of the electron and its speed v: K.E = \(\frac{1}{2}\)mv\(^2\)
K.E = eV\(_0\) = \(\frac{1}{2}\)mv\(^2\)
v = \(\sqrt{\frac{2eV_o}{m}}\)
Substituting v = \(\sqrt{\frac{2eV_o}{m}}\) into the de Broglie equation
\(\lambda\) = \(\frac{\text{h}}{m \times \sqrt{\frac{2eV_o}{m}}}\) = \(\frac{\text{h}}{\sqrt{\frac{2eV_o m^2}{m}}}\)
\(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\)
Note: use the fact that \(\sqrt{m}\) x \(\sqrt{m}\) = m
Source: WAEC / JAMB / NECO Classroom Archive (Record #72089)
Physics
UTME / JAMB 2024
Reflection of Light
Solution
A light ray passing from air into water at an angle of 30º from the normal in air would
A
make an angle greater than 30\(^0\) in water surface
B
be completely absorbed by the water surface
C
be completely reflected by the water surface
D
make an angle less than 30\(^0\) from the normal in water
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Observations show that when the light travels from a medium A(air) to a more dense medium B(water), the refracted ray bends towards the normal and the angle of refraction (r) is, therefore, less than the angle of incidence. vice versa.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70362)
Physics
UTME / JAMB 2024
Reflection of Light
Solution
In a solar panel, solar beam is concentrated by using
A
Concave mirror
B
Convex mirror
C
Plane mirror
D
Triangular mirror
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Concave mirrors are designed to focus sunlight onto a specific area, enhancing the intensity of the solar energy. This concentrated energy can then be utilized for heating or converted into electricity in solar thermal and photovoltaic systems.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70369)
Physics
UTME / JAMB 2024
Reflection of Light
Solution
Which of the following is the best as shaving mirror?
A
Concave mirror
B
Convex mirror
C
plane mirror
D
parabolic mirror
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Concave Mirror: A concave mirror can focus light rays, allowing for a magnified image of the face. This magnification helps in seeing details more clearly, which is particularly useful for shaving.
Convex Mirror: A convex mirror creates a smaller, wider image, which is not suitable for precise tasks like shaving.
Plane Mirror: A plane mirror shows a true reflection but does not magnify, making it less effective for grooming.
Parabolic Mirror: Parabolic mirrors are used for specific applications, like telescopes, but not for shaving.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70430)
Physics
UTME / JAMB 2024
Reflection of Light
Solution
A red shirt under a red light appears pale because red
A
reflects nearly all light falling on it
B
Unable to reflect lights falling on it
C
absorbs all lights falling on it
D
absorbs other colours and reflects red
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Under red light, the shirt reflects the red light but does not reflect other colors, which makes it look less vibrant or pale, compared to being illuminated by white light that contains all colors.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70479)
Physics
UTME / JAMB 2024
Reflection of Light
Solution
A practical application of total internal reflection is found in
A
rainbow
B
diverging lens
C
converging lens
D
fiber optics
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Fiber optics:
Fiber optics: Utilizes total internal reflection to transmit light through optical fibers, allowing for efficient data transfer.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70524)