MDCAT Physics chapter 14: Physics · Dawn of modern physics MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
20 exam-style questions on Physics · Dawn of modern physics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Ten Physics · Dawn of modern physics questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. For a given metal, the maximum kinetic energy of the photoelectrons (y-axis) is plotted against the frequency of the incident light (x-axis), giving a straight line. The slope of this line is
A
the ratio h/e
B
Planck's constant h
C
the work function of the metal
D
the threshold frequency of the metal
Show answer
Answer: B. K.E.max = hf − φ, a straight line of slope h. It cuts the frequency axis at the threshold frequency f0, and its extension meets the energy axis at −φ. The ratio h/e is the slope of a graph of stopping potential against frequency.
2. Light of a frequency above the threshold falls on a metal surface. If the frequency of the light is doubled, the maximum kinetic energy of the photoelectrons
A
becomes exactly twice as large
B
increases, but by less than a factor of two
C
becomes more than twice as large
D
does not change at all
Show answer
Answer: C. K.E.max = hf − φ. At frequency 2f it becomes 2hf − φ = 2(hf − φ) + φ, which is more than twice the old value by φ. It would exactly double only if the work function were zero.
3. No photoelectrons are emitted from a metal surface, however intense the light, if
A
the wavelength of the light is shorter than the threshold wavelength
B
the frequency of the light is below the threshold frequency
C
the light falls on the surface for less than a few seconds
D
the intensity of the light is below a certain minimum value
Show answer
Answer: B. Each electron absorbs a single photon, and emission needs hf ≥ φ, the work function. Below the threshold frequency no photon has enough energy, so brighter light (more photons) does not help; when emission does occur, it starts without any delay.
4. In Compton scattering of X-rays by free electrons, the increase in wavelength of the scattered photons is greatest for a scattering angle of
A
90°
B
0°
C
45°
D
180°
Show answer
Answer: D. Δλ = (h/m0c)(1 − cos θ), which is largest when cos θ = −1, that is θ = 180° (photon scattered straight back), giving 2h/m0c. At 0° there is no shift at all.
5. A rod of proper length 10 m moves along its own length at 0.6c past an observer. The length that the observer measures is
A
12.5 m
B
6.4 m
C
8.0 m
D
10 m
Show answer
Answer: C. l = l0√(1 − v2/c2) = 10 × √(1 − 0.36) = 10 × 0.8 = 8.0 m. 12.5 m divides by the factor instead of multiplying, and 6.4 m forgets the square root.
6. A lamp emits 6.6 W of light, all of wavelength 600 nm. Taking h = 6.6 × 10−34 J s and c = 3.0 × 108 m s−1, the number of photons it emits per second is
A
2.0 × 1019
B
2.0 × 1018
C
3.3 × 10−19
D
5.0 × 10−20
Show answer
Answer: A. One photon has E = hc/λ = (6.6 × 10−34 × 3.0 × 108)/(6.0 × 10−7) = 3.3 × 10−19 J, so the number per second = 6.6 ÷ (3.3 × 10−19) = 2.0 × 1019. Taking 600 nm as 6.0 × 10−8 m gives 2.0 × 1018, and 3.3 × 10−19 is the energy of one photon in joules.
7. An electron microscope can show much finer detail than an optical microscope mainly because
A
the electrons have a much greater wavelength than visible light
B
the electrons have a much shorter wavelength than visible light
C
electrons travel faster than light inside the microscope
D
the specimen gives out its own light when struck by electrons
Show answer
Answer: B. The smallest detail a microscope can resolve is set by the wavelength used. Electrons accelerated through a few thousand volts have de Broglie wavelengths far shorter than the 400–700 nm of visible light, so much finer detail can be resolved.
8. Davisson and Germer confirmed the wave nature of electrons by observing
A
the emission of electrons from a metal lit by ultraviolet light
B
the scattering of X-rays by free electrons in graphite
C
the diffraction of an electron beam by a nickel crystal
D
the deflection of α-particles by a thin gold foil
Show answer
Answer: C. Davisson and Germer found that electrons scattered from a nickel crystal show diffraction maxima at angles that fit λ = h/mv, showing that moving particles behave as waves. X-ray scattering by electrons is the Compton effect, and α-particle scattering led to the nuclear model of the atom.
9. In a nuclear reaction the products have 2.0 × 10−7 kg less mass than the reactants. Taking c = 3.0 × 108 m s−1, the energy released is
A
60 J
B
1.8 × 1013 J
C
9.0 × 109 J
D
1.8 × 1010 J
Show answer
Answer: D. E = Δmc2 = 2.0 × 10−7 × (3.0 × 108)2 = 1.8 × 1010 J. 60 J is mc (c not squared), and 9.0 × 109 J wrongly uses ½mc2.
10. According to the special theory of relativity, the speed of light in free space
A
depends on the speed of the source of light
B
is greater for an observer moving towards the source
C
is the same only for observers at rest on the earth
D
is the same for all observers in inertial frames
Show answer
Answer: D. This is the second postulate of special relativity: c is the same in all inertial frames, whatever the motion of the source or the observer. Time dilation and length contraction follow from it.
Physics · Dawn of modern physics: questions students ask
How many Physics · Dawn of modern physics MCQs are there?
20, each with the answer and a short explanation, across 2 topics: Photons and the photoelectric effect, Wave nature of particles. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Physics chapter 14 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
