phy al 2010 mc question

用戶31980

2013-04-04 7:36 am

can somebody show me the steps or give me the 題解 of this question?
40. An a-particle (4/2He) of initial kinetic energy 7.7 MeV approaches a gold nucleus 197/79Au) from far away.
The a-particle is deflected as shown(sorry for not providing the figure...). Estimate the kinetic energy of the a-particle at point P(which is the closest point between the a-particle and the gold nucleus)where it is 6.5 x 10=14 m from the gold nucleus which is assumed to be stationary throughout.
Given: charge of an electron 1.60 x 10^-19 C
permittivity of free space = 8.85 x 10=12 F m^-1

A. 3.5 MeV
B. 4.2 MeV
C. 7.3 MeV
D. 11.2 MeV

回答 (1)

天同

2013-04-04 8:09 am

✔ 最佳答案

Electrostatic potential energy of alpha particle at P, V = kqQ/r
where k is the electrostatic constant ( = 9 x 10^9 N.m^2/C^2)
q is the charge of an alpha particle (= 2e, where e is the electronic charge)
Q is the charge of a gold nucleus (=79e)
r is the distance between the alpha particle and the gold nucleus

Hence, V = (9x10^9).(2e).(79e)/6.5x10^-14 J = 5.6x10^-13 J = 3.5 MeV
By conservation of energy,
7.7 = KE + 3.5
where KE is the kinetic energy of the alpha particle at P
hence, KE = (7.7 - 3.5) MeV = 4.2 MeV

The answer is option B

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