Two vertical metal plates are separated by a distance d in a vacuum, as shown in Fig. 7.1. [Figure 7.1] The potential difference (p.d.) between the plates is V. A nucleus with charge +q is initially at rest on plate X. The nucleus is accelerated by the uniform electric field from plate X along a horizontal path to plate Y.
- (a)(i)State expressions, in terms of some or all of d, q and V, for: the magnitude of the electric field strength electric field strength = ........................................................... [1][1]
- (a)(ii)the magnitude of the electric force acting on the nucleus force = ..................................................................................... [1][1]
- (a)(iii)the kinetic energy of the nucleus when it reaches plate Y. kinetic energy = ........................................................... [1][1]
- (b)(i)State the change, if any, in the kinetic energy of the nucleus on reaching plate Y when the following separate changes are made. The distance d is halved, but the p.d. V remains the same. [1][1]
- (b)(ii)The nucleus is replaced by a different nucleus that is an isotope of the original nucleus with fewer neutrons. [1][1]
- (c)(i)The nucleus is carbon-14 (¹⁴₆C). This nucleus decays to form a new nucleus by releasing a β⁻ particle and only one other particle of negligible mass. (i) Calculate the nucleon number and the proton number of the new nucleus. nucleon number = ........................................................... proton number = ........................................................... [1][1]
- (c)(ii)State the name of the particle of negligible mass. [1][1]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [1] electric field strength = V/d
- [1] force = Vq/d
- [1] kinetic energy = Vq
- [1] no change
- [1] no change
- [1] nucleon number = 14 and proton number = 7
- [1] (electron) antineutrino
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