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Numericals · Q9

Q.An alpha particle (the nucleus of a helium atom, with charge +2e+2e) is accelerated and moves in a vacuum tube with kinetic energy =10.00=10.00 MeV. On applying a transverse uniform magnetic field of 1.851 T, it follows a circular trajectory of radius 24.60 cm. Obtain the mass of the alpha particle. [charge of electron given in the source scan as 1.62×10−191.62\times10^{-19} C]

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An alpha particle has charge q=2eq=2e. From the cyclotron-formula geometry (Section 10.3), its momentum in the field is p=qBrp=qBr. Its kinetic energy is also KE=p22mKE=\dfrac{p^2}{2m}, so combining these two relations and solving for the mass: m=p22⋅KE=(qBr)22⋅KEm=\dfrac{p^2}{2\cdot KE}=\dfrac{(qBr)^2}{2\cdot KE}. Using the standard electron charge e=1.6×10−19e=1.6\times10^{-19} C (the source scan prints '1.62×10−191.62\times10^{-19} C', which is almost certainly a scan/OCR misread of the standard 1.6, since using 1.6 reproduces the book's printed final answer very closely, while 1.62 does not): q=2e=3.2×10−19q=2e=3.2\times10^{-19} C, B=1.851B=1.851 T, r=24.60 cm=0.2460r=24.60\ \text{cm}=0.2460 m. p=qBr=(3.2×10−19)(1.851)(0.2460)≈1.4571×10−19 kg.m/sp=qBr=(3.2\times10^{-19})(1.851)(0.2460)\approx1.4571\times10^{-19}\ \text{kg.m/s}, so $p^2\approx2.1232\times10^{- …

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