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प्रश्न
An alpha particle (the nucleus of helium atom) (with charge +2e) is accelerated and moves in a vacuum tube with kinetic energy = 10.00 MeV. On applying a transverse a 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 = 1.62 × 10-19 C]
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उत्तर
Data: 1 eV = 1.62 × 10-19 J,
E = 10 MeV = 107 × 1.62 × 10-19 J = 1.62 × 10-12 J,
B = 1.851 T,
r = 0.246 m,
e = 1.62 × 10-19 C
Charge of an α-particle,
q = 2e = 2(1.62 x 10-19) = 3.24 x 10-19 C
r = `("m"_alpha "v")^2/("qB")` and E = `1/2 "m"_alpha "v"^2`
∴ `"r"^2 = ("m"_alpha "v")^2/("q"^2"B"^2)` and `2"Em"_alpha = ("m"_alpha"v")^2`
∴ `"r"^2 = (2"Em"_alpha)/("q"^2"B"^2)`
∴ `"m"_alpha = ("qBr")^2/"2E"`
`= [(3.24 xx 10^-19)(1.851)(0.246)]^2/(2(1.62 xx 10^-12))`
= (3.24 × 10-26)(1.851 × 0.246)2
= (3.24 × 10-26)(0.455)2
= (3.24 × 10-26)(0.2070)
= 6.7068 × 10-27 kg
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