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Nuclear Energy Strategy announced at CNA2026
At the Canadian Nuclear Association Conference (CNA2026) in Ottawa, Ontario, on April 29, Minister of Energy and Natural Resources Tim Hodgson announced that Natural Resources Canada (NRCan) is developing a new Nuclear Energy Strategy for the country. The strategy, which is slated to be released by the end of this year, will be based on four objectives: 1) enabling new nuclear builds across Canada, 2) being a global supplier and exporter of nuclear technology and services, 3) expanding uranium production and nuclear fuel opportunities, and 4) developing new Canadian nuclear innovations, including in both fission and fusion technologies.
F. C. Schoenig, K. S. Quisenberry, D. P. Stricos, and H. Bernatowicz
Nuclear Science and Engineering | Volume 26 | Number 3 | November 1966 | Pages 393-398
Technical Paper | doi.org/10.13182/NSE66-A17362
Articles are hosted by Taylor and Francis Online.
The temperature dependence of the thorium-oxide resonance integral has been measured over a wide (20 to 1550 °C) temperature range. The activation method was used; the 310 keV γ ray from the decay of 233Pa was measured with a multichannel pulse-height analyzer. Measurements were performed on ThO2 rods of 0.490− and 0.353−in. diam. (surface-to-mass ratio = 0.340 and 0.465 cm2/g, respectively). The temperature dependence of the thorium-oxide resonance integral was found not to be a linear function of either (t − t0) or (√T − √T0), where t and T and centigrade and Kelvin temperature, and t0 and T0 are 20°C, and 293°K, respectively. Thus the familiar forms of the temperature dependence of the effective resonance integral, namely RI(T)/RI(T0) = 1 + α (t − t0) = 1 + β × (√T − √To) are not appropriate representations of the data. The Doppler coefficient in a 1/E spectrum is defined by α0 = [1/RI(T)] [dRI(T)/ dT] where RI(T) is the effective resonance integral of the sample excluding the 1/v contribution, and T is the temperature of the sample. It has been found that α0 = [(0.16 ± 0.01)/T] yields a good fit to the experimental data of both sample sizes. It follows that RI(T) = RI(T0) (T/T0)(0.16 ± 0.01).