question 1

Question

A)

 

A proton is fired from far away toward the nucleus of a mercury atom. Mercury is element number 80, and the diameter of the nucleus is 14.0 fm.

If the proton is fired at a speed of 3.2×10^7 m/s , what is its closest approach to the surface of the nucleus? Assume the nucleus remains at rest.

 

 

B)

 

 

The electric field strength is 2.40×10^4 N/C inside a parallel-plate capacitor with a 1.00 mm spacing. An electron is released from rest at the negative plate.

 

What is the electron's speed when it reaches the positive plate?

Answer 1

Hi Moe!

 

(a)  Considering this only as an electric force problem, we can use conservation of energy (since the electric force is conservative).

 

The value of electric potential energy is defined as:

 

U_E = kq₁q₂/r

 

where r is the distance between the charges (and we can keep the positive and negative signs on positive and negative charges).

 

When the proton is very, very far from the mercury nucleus, we have r --> ∞, meaning:

 

(U_E)_far= 0 (effectively)

 

At the far position, the proton is launched with speed 3.2 x 10⁷ m/s.  This is about one tenth the speed of light, which is a decent threshold for ignoring relativistic effects on the energy.  Thus, the kinetic energy is:

 

KE_far = (1/2)m_pv²

 

where

 

m_p = mass of proton = 1.67x10^-27 kg

v = 3.2 x 10⁷ m/s

 

So, far from the mercury nucleus, the total energy of the proton is:

 

E_far = (U_E)_far + KE_far = 0 + KE_far = (1/2)m_pv²

 

When the proton is at its closest approach to the mercury nucleus, if it is fired directly along a line, it will momentarily come to rest (just like a ball tossed directly up in the air will momentarily come to rest at its highest point).  At that point:

 

(U_E)_close = kq₁q₂/r_close

 

where

 

q₁ = proton charge = e (fundamental charge, e = 1.6x10^-19 C)

q₂ = mercury nucleus charge = Ze, where Z = atomic number of mercury (Z = 80, stated in the problem)

 

So 

 

(U_E)_close = k(e)(80e)/r_close

 

At this point, the proton is momentarily at rest, so its kinetic energy is zero.  So its total energy near is:

 

E_close = (U_E)_close + KE_close = k(e)(80e)/r_close + 0 = 80ke²/r_close

 

By using conservation of energy to equate E_far and E_close, you make an equation with only one unknown:  r_close

 

E_far = E_close

(1/2)m_pv² = 80ke²/r_close

 

Everything except r_close in this equation is either given in the problem or is a universal constant you can look up.  From here, you can solve for r_close.

 

Now, for a uniform sphere of charge (which we can presume the nucleus to be), this distance (r_close) is measured from the proton (which we can assume is a point mass) to the center of the nucleus sphere.  Hence, the distance of the proton from the surface of the nucleus at closest approach is r_close MINUS the radius of the nucleus, which you can calculate from given information.

 

(b)  There are a couple ways to do this one, but since (a) involved energy (and work, in a way), we can do this one that way, too.

 

For a uniform electric field (such as we usually assume between the two plates of a capacitor), there is a direct relationship between electric potential and distance traveled along the field.  The electric field points from positive charges (high potential) to negative charges (low potential).  Thus, as you travel along the field in the direction it points, you go down in potential.  If you travel along the field line opposite the way it points, you go up in potential.

 

The relationship for a uniform electric field between potential (or potential difference) and distance traveled along the field line between points a and b is:

 

V_ab = Ed

 

where V_ab is the potential change between points a and b, E is the electric field strength, and d is the d is the distance from a to b.  With this, we can calculate the potential difference between the plates.

 

V_ab = (2.4 x 10⁴ N/C)(0.001 m)

 

The change in kinetic energy of the electron during this motion is equal to the net work done on it, and only the electric field if doing work (positive work, since the electron is moving the way it naturally "wants" to go).  The value of the change in kinetic energy is:

 

W = ΔKE = KE_f - KE_o = qV_ab

 

In this case, q = e, and V_ab was calculated above.  So a value for KE_f - KE_o can be determined, and since the electron starts from rest (KE_o = 0), this value equals KE_f.  Then:

 

KE_f = (1/2)m_ev²

 

where

 

m_e = mass of the electron

v = electron's final speed 

 

(this assumes again that relativity need not be considered here; that would change the form of the KE expression, if it were)

 

I hope this helps!  Let me know if you have any other questions, or would like to check an answer.