A-LevelPhysicsNuclear physics9702/22 · Feb/Mar 2017 · Q78 Marks
A nucleus of bismuth-212 (²¹²Bi) decays by the emission of an α-particle and γ-radiation.
- (a)State the number of protons and the number of neutrons in the nucleus of bismuth-212.[1]
- (b)The γ-radiation emitted from the nucleus has a wavelength of 3.8 pm. Calculate the frequency of this radiation.[3]
- (c)Explain how a single beam of α-particles and γ-radiation may be separated into a beam of α-particles and a beam of γ-radiation.[2]
- (d)The α-particle emitted from the bismuth nucleus has an initial kinetic energy of 9.3 × 10⁻¹³ J. As the α-particle moves through air it causes the removal of electrons from atoms. The α-particle loses energy and is stopped after removing 1.8 × 10⁵ electrons as it moved through the air. Determine the energy, in eV, needed to remove one electron.[2]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.
- [A1] (a) number of protons = 83 and number of neutrons = 129
- [C1] (b) λ = 3.8 × 10⁻¹²
- [C1] (b) f = 3.0 × 10⁸ / 3.8 × 10⁻¹²
- [A1] (b) f = 7.9 × 10¹⁹ (7.89 × 10¹⁹) Hz
- [M1] (c) use an electric field (at an angle to the beam)
- [A1] (c) α is deflected and γ is undeflected
- [C1] (d) energy = 9.3 × 10⁻¹³ / 1.8 × 10⁵ (= 5.17 × 10⁻¹⁸ J)
- [A1] (d) = 5.17 × 10⁻¹⁸ / 1.6 × 10⁻¹⁹ = 32 (32.3) eV
- [C1] (d) OR energy = 9.3 × 10⁻¹³ / 1.6 × 10⁻¹⁹ (= 5.81 × 10⁶ eV)
- [A1] (d) OR = 5.81 × 10⁶ / 1.8 × 10⁵ = 32 (32.3) eV
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