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100+ Free Sachsen-Anhalt Abitur Physics Practice Questions

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2026 Statistics

Key Facts: Sachsen-Anhalt Abitur Physics Exam

4 Themenbereiche

Einführungsphase (Optik, Radioaktivität, Gravitation, Klimaphysik) plus Qualifikationsphase Themenbereiche Schwingungen und Wellen, Elektrodynamik und Quantenphysik

Fachlehrplan Physik Gymnasium Sachsen-Anhalt, Stand 01.08.2022

4 Aufgaben, 3 bearbeiten

Since the 2025 Abitur, candidates choose 3 of 4 offered written Aufgaben

Hinweise für die schriftliche Abiturprüfung ab 2025 Physik, Ministerium für Bildung Sachsen-Anhalt

300 / 255 min

Written Bearbeitungszeit including Auswahlzeit at erhöhtem (120 BE) und grundlegendem (90 BE) Anforderungsniveau

Hinweise für die schriftliche Abiturprüfung ab 2025 Physik, Ministerium für Bildung Sachsen-Anhalt

23.04.2026

Written Physik date in the 2026 main session; Nachtermin 20.05.2026

Ministerium für Bildung des Landes Sachsen-Anhalt, Abiturtermine 2026

0–15 Punkte

Grading scale for each Prüfungsfach, from Note 1 (15–13 Punkte) to Note 6 (0 Punkte)

Verordnung über die gymnasiale Oberstufe, Sachsen-Anhalt

200 / 100 Punkte

Minimum Gesamtqualifikation: 200+ from Block I (§38) and 100+ from the five Prüfungselemente in Block II (§39), out of 900 maximum

Verordnung über die gymnasiale Oberstufe, Sachsen-Anhalt

Free 100-question English-language MCQ study bank for Sachsen-Anhalt Abitur Physik, mapped to the Fachlehrplan's Einführungsphase (Optik, Radioaktivität, Gravitation, Klimaphysik) and Qualifikationsphase Themenbereiche (Schwingungen und Wellen, Elektrodynamik, Quantenphysik). Over 60 worked-calculation items. Official exam: German written Klausur, 4 Aufgaben choose 3, roughly 300 min (eA) / 255 min (gA) including Auswahlzeit; written date 23.04.2026. Not an official-format simulation; no fee for regular school candidates.

Sample Sachsen-Anhalt Abitur Physics Practice Questions

Try these sample questions to test your Sachsen-Anhalt Abitur Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A ray of light travels from air (n = 1.00) into glass (n = 1.50) and strikes the boundary at an angle of incidence of 40°. Using Snell's law, what is the angle of refraction inside the glass?
A.25.4°
B.74.6°
C.26.7°
D.60.0°
Explanation: Snell's law states n1·sinθ1 = n2·sinθ2. Solving for θ2: sinθ2 = (n1/n2)·sinθ1 = (1.00/1.50)×sin40° ≈ 0.4285, so θ2 = sin⁻¹(0.4285) ≈ 25.4°. Because light enters the optically denser medium, it bends toward the normal, so θ2 is smaller than θ1.
2An object is placed 30 cm in front of a thin converging lens with a focal length of 10 cm. Using the thin-lens equation 1/f = 1/a + 1/b, at what distance b behind the lens does the real image form?
A.15 cm
B.7.5 cm
C.20 cm
D.30 cm
Explanation: Rearranging 1/f = 1/a + 1/b gives 1/b = 1/f − 1/a = 1/10 − 1/30 = 3/30 − 1/30 = 2/30 = 1/15, so b = 15 cm. Because the object distance a exceeds f, this is a real, inverted image formed on the far side of the lens.
3Light traveling inside glass (n = 1.50) strikes the glass–air boundary. What is the critical angle beyond which total internal reflection occurs?
A.41.8°
B.48.2°
C.26.4°
D.20.9°
Explanation: Total internal reflection begins at the critical angle where the refracted ray grazes the boundary at 90°. From Snell's law, n·sinθc = 1·sin90°, so sinθc = 1/n = 1/1.50 ≈ 0.667, giving θc = sin⁻¹(0.667) ≈ 41.8°.
4Why does the simple ray (Strahlenmodell) model of light fail to fully explain the dispersion of white light into a spectrum by a glass prism?
A.Because the refractive index of glass depends on the wavelength of light, a fact the basic ray model does not include
B.Because light travels faster in glass than in air
C.Because red light is absorbed more strongly by glass than blue light
D.Because the ray model assumes light is a particle, and particles cannot be refracted
Explanation: Dispersion arises because the refractive index n of a medium depends on wavelength — a property of the material itself. The purely geometric ray model treats n as a single constant per material, so on its own it cannot predict that different colours bend by different amounts; that requires incorporating the wavelength-dependence of n.
5A radioactive sample has an initial activity of 800 Bq and a half-life of 8 days. What is its activity after 24 days?
A.100 Bq
B.266.7 Bq
C.200 Bq
D.400 Bq
Explanation: 24 days corresponds to 24/8 = 3 half-lives. Each half-life halves the activity in succession: 800 → 400 → 200 → 100 Bq. In general A(t) = A0·(1/2)^(t/T½) = 800·(1/2)³ = 100 Bq.
6Carbon-14 has a half-life of 5730 years. Using λ = ln(2)/T½, what is its decay constant λ?
A.1.21 × 10⁻⁴ yr⁻¹
B.1.75 × 10⁻⁴ yr⁻¹
C.5.25 × 10⁻⁵ yr⁻¹
D.8270 yr⁻¹
Explanation: λ = ln(2)/T½ = 0.6931/5730 yr ≈ 1.21×10⁻⁴ yr⁻¹. This constant gives the probability per unit time that any one nucleus decays, and underlies the exponential decay law N(t) = N0·e^(−λt) used in C-14 dating.
7Uranium-238 (₉₂²³⁸U) undergoes alpha decay. Which nuclide is produced?
A.Thorium-234 (Th-234)
B.Protactinium-234 (Pa-234)
C.Thorium-238 (Th-238)
D.Uranium-234 (U-234)
Explanation: An alpha particle is a helium-4 nucleus (₂⁴He), so alpha decay reduces the mass number by 4 and the atomic number by 2: A = 238−4 = 234, Z = 92−2 = 90, which is thorium. So ₉₂²³⁸U → ₉₀²³⁴Th + ₂⁴He.
8What happens at the nuclear level during beta-minus (β⁻) decay?
A.A neutron in the nucleus converts into a proton, emitting an electron and an antineutrino
B.A proton converts into a neutron, emitting a positron
C.The nucleus emits a helium nucleus consisting of two protons and two neutrons
D.The nucleus emits a high-energy photon without changing its proton or neutron number
Explanation: In β⁻ decay, a neutron transforms into a proton via the weak interaction, releasing an electron (the 'beta particle') and an antineutrino. Because a neutron becomes a proton, the atomic number Z increases by 1 while the mass number A stays the same, moving the element one place to the right in the periodic table.
9Which property of ionizing radiation is exploited by a Geiger–Müller counter to detect individual decay events?
A.Radiation ionizes gas atoms inside the tube, and the resulting ions and electrons are accelerated by a high voltage to trigger a measurable current pulse
B.Radiation heats the counting gas, and the resulting pressure change is measured directly
C.Radiation is absorbed by a fluorescent screen and directly converted into a digital readout without any electrical signal
D.Radiation changes the colour of a chemical strip inside the tube, which is read optically
Explanation: A Geiger–Müller tube contains a low-pressure gas between two electrodes at high voltage. An ionizing particle creates ion pairs along its path; the strong field accelerates these charges, causing an avalanche that produces a detectable current pulse — one pulse per ionizing event.
10Using Newton's law of gravitation F = γ·m1·m2/r² with γ = 6.674×10⁻¹¹ N·m²/kg², what is the gravitational force between two 1000 kg masses whose centres are 2 m apart?
A.1.67 × 10⁻⁵ N
B.3.34 × 10⁻⁵ N
C.3.34 × 10⁻⁸ N
D.2.50 × 10⁵ N
Explanation: F = γ·m1·m2/r² = 6.674×10⁻¹¹ × (1000×1000) / 2² = 6.674×10⁻¹¹ × 10⁶ / 4 ≈ 1.67×10⁻⁵ N. Gravitational force between everyday masses is extremely small, which is why it is never noticeable compared with Earth's pull on those same objects.

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