# OCR A level physics - Nuclear physics module

A polonium-210 atom has mass 3.5x10^-25kg and each decay from one of these atoms produces an energy of 5.3MeV. a) How many decays per second are required to supply a power of 1000W? b) What mass of polonium-210 is required for this power? The decay constant of polonium-210 is 5.8x10^-8

Answers

1000 W is 1000 J per sec. Part (a) is asking how many packets of 5.3MeV there are in 1000 J. 1eV=1.6x10^-19 J.The decay constant is the probability of an individual nucleus decaying per second. You now know how many nuclei per second are in fact decaying, so you can find out how many nuclei there must be altogether in your lump of polonium. You know the mass of a polonium atom so you can find the mass of the lump.

03 January 2015

Start by considering what power is? Energy transferred per unit time, thus 1000 W=1000 J/sNow we need how much energy each decay gives= 5.3MeV=5.3*10^6 eV=5.3*10^6*1.6*10^-19 J from the definition of eV.Then we ask how many decays go into 1000W? Well that's 1000/(5.3*10^6*1.9*10-19) which is about 10^15. Rather a lot!!

28 January 2015

28 January 2015

1000w=1000j/s 5.3ev=5.3×1.6×10^-19j so no of decay=1000/(5.3×1.6×10^_19)= mass of polanium=(210×no of decay)/5.8×10^-8 amu

23 March 2015

First convert 5.3 MeV to Joules by multiplying by 1.602*10^-19 this then equals the energy in joules per emitted particle. You are told you need to fulfil a power requirement if 1000W which is the same as 1000 J every second so to find the number of decays per second you should divide 1000 by 5.3*1.602*10^-19For the second question use the radioactive decay law. You know the number of disintegration so from the first calculation, call this n. Let the number of original atoms in the sample to produce this power be N. After 1 second therefore you have the original number of atoms minus the disintegrations or N-n. By the radioactive decay lawN-n = N*exp^(-lamda*t) but t = 1sec, lambda is decay constant givenRearranged you get N =n/ {1 - exp^(-lambda)}You can calculate N the original number of atoms of polonium required to produce this power as you know n from first answer and you are given lamdaCalculation of the mass of Polonium 210 required is then simply a matter of calculating the product if N and the mass of one Polonium atom which is given

29 August 2015

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