It depends on the half life of the element in question. The most comparable concrete thing we can compare this to with real numbers because we know it works is an RTG. RTGs are solid-state generators, but people could colloquially refer to them as “batteries” and not be terribly wrong. They take a quantity of a radioactive material and allow it to decay, using the heat given off to establish a thermal gradient which is then converted to electricity via thermocouples. Most of these are “fueled” with Pu-238 (at least the ones for spacecraft), which has a half life of 87.7 years. That means in 87.7 years, if you started with 4kg of Pu when you built it, you’d have only 2kg of Plutonium left. If the Pu decayed only into stable isotopes (it doesn’t) then your radioactive emissions/decay would also be exactly halved at this time. If the electrical system is perfectly efficient this would also halve the electrical power produced.

I provide this all as background because to answer your question you have to know three key factors about the device to determine the lifetime of the battery. The half-life of the isotope used, the minimum electrical requirements of the device you’re powering, and the amount of radioactive material in the initial battery. The battery’s lifetime is determined by when decay will decrease the ongoing energy output below the minimum current and voltage requirements needed by the battery. The longer the half life of the isotope, the slower this decrease is and the less initial overpowering that is required.

Ex. If you use an isotope with a 12.5 year half life for a “50-year” battery, you would need to start with 8 times the material needed for your minimum power output requirements. If you use an isotope with a 200 year half life, you only need 19% more starting mass than you minimum requirement. The first battery will produce 8x the power at the very beginning, while the second will only produce 18% more.