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A proton is launched from very far away directly toward the nucleus of a Mercury atom. Ifthe initial speed of the proton is 4.0×107m/s, how close to the nucleus does it get? Take theradius of a Mercury nucleus to be 7.0 fm and assume that it remains at rest.

A proton is launched from very far away directly toward the nucleus of a Mercury atom. Ifthe initial speed of the proton is 4.0×107m/s, how close to the nucleus does it get? Take theradius of a Mercury nucleus to be 7.0 fm and assume that it remains at rest.

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A proton is launched from very far away directly toward the nucleus of a Mercury atom. Ifthe initial speed of the proton is 4.0×107m/s, how close to the nucleus does it get? Take theradius of a Mercury nucleus to be 7.0 fm and assume that it remains at rest.

I have tried to answer the question above, and my solution is attached. I was just wondering if the solution that I have is right? I am a bit confused about whether it would be an acceleration or a deceleration?

 

We can treat the nucleus and the proton like point charges. The atomic number of Mercury is Z = 80 so the charge of the nucleus is Q = Ze. The charge of the proton is q = e. The force on the proton as it moves toward the nucleus is given by Coulomb's law and has the magnitude Ze? F 4TEor2 (1) This force is directed away from the nucleus. Therefore, the acceleration of the proton has a magnitude a = F/m giving Ze? a = (2) 4TEomr2 also directed away from the nucleus. It is a deceleration. The proton will reach its closest approach to the nucleus when v = 0. We can use one of the kinematic equations, v? = v3 + 2ax, to get the final speed of the proton where we'll use r instead of r with the magnitude of a given by (2). Thus, we have Ze? v² = vổ - (3) 2TEomr We have used a minus sign because it's decelerated motion. Setting v = 0 and solving for r gives us the distance of closest approach Ze? r = (4) The result becomes (80)(1.602 × 10-19C)² = 13.8 fm (5) r = 27(8.85 x 10-12C²/N-m)(1.67 x 10–27kg)(4.0 × 107m/s)2 This is the distance from the center of the nucleus. Because the radius of the nucleus is 7.0 fm, the closest approach to the nucleus is 6.8 fm.
Transcribed Image Text:We can treat the nucleus and the proton like point charges. The atomic number of Mercury is Z = 80 so the charge of the nucleus is Q = Ze. The charge of the proton is q = e. The force on the proton as it moves toward the nucleus is given by Coulomb's law and has the magnitude Ze? F 4TEor2 (1) This force is directed away from the nucleus. Therefore, the acceleration of the proton has a magnitude a = F/m giving Ze? a = (2) 4TEomr2 also directed away from the nucleus. It is a deceleration. The proton will reach its closest approach to the nucleus when v = 0. We can use one of the kinematic equations, v? = v3 + 2ax, to get the final speed of the proton where we'll use r instead of r with the magnitude of a given by (2). Thus, we have Ze? v² = vổ - (3) 2TEomr We have used a minus sign because it's decelerated motion. Setting v = 0 and solving for r gives us the distance of closest approach Ze? r = (4) The result becomes (80)(1.602 × 10-19C)² = 13.8 fm (5) r = 27(8.85 x 10-12C²/N-m)(1.67 x 10–27kg)(4.0 × 107m/s)2 This is the distance from the center of the nucleus. Because the radius of the nucleus is 7.0 fm, the closest approach to the nucleus is 6.8 fm.
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