Alpha particles which have a charge +2e and mass 6.64 ✕ 10−27 kg are initially at rest and are fired directly at a stationary lead nucleus (charge +82e) at a speed of 1.05 ✕ 107 m/s. Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the lead nucleus remains stationary. Assume the alpha particles are initially very far from a stationary lead nucleus.

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A: It is given that

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A: Thank you

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A: According to Einstein relation,  E=∆mc2 Change in mass is given by,  ∆m=3.0221+4.0834-6.1450+0.0087…

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A: Given,

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A: Given : r = 0.5 Å q1 = -e (for electron charge is negative) q2 = 6e 1Å (Angstrom)= 10^-10m…

Q: For the following reaction, what is the energy released, in GJ/mol? (1 GJ = 1E9 J) Use 2.998E8 m/s…

A: Given,

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A: given ,  mass of the proton , m = 1.672 x 10-27 Kg velocity of the proton , v = 3.0 x 106 m/s

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A: The force acquired by the particles is F=kq1q2r2=9×109 N·m2/C23.20×10-19 C1.36×10-17 C0.052…

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A: Given Alpha particle q1          = 2e Gold atom protons q2 = 79 e Gold atom neutrons       = 118…

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A: given that,speed of electron, v=6.8×105ms

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A: Given: Accelerating voltage = 1.5 kV Mass of electron = 9.11 X 10-31 kg

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.