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beks73 [17]
3 years ago
5

During a famous experiment in 1919, Ernest Rutherford shot doubly ionized helium nuclei (also known as alpha particles) at a gol

d foil. He discovered that virtually all of the mass of an atom resides in an extremely compact nucleus. Suppose that during such an experiment, an alpha particle far from the foil has a kinetic energy of 6.0 MeV. If the alpha particle is aimed directly at the gold nucleus, and the only force acting on it is the electric force of repulsion exerted on it by the gold nucleus, how close will it approach the gold nucleus before turning back
Physics
2 answers:
Delvig [45]3 years ago
5 0

Answer:

The answer to the question is;

The alpha particle will approach the gold nucleus up to 3.79 × 10⁻¹⁴ m before turning back.

Explanation:

To solve the question, we note that

There are  79 electrons the gold tom and 2 electrons in the alpha particle which is the helium nuclear

We use the energy conservation principle to obtain

W_{net} = \int\limits^{r_{min}}_{inf} {F_e} \, dr = Δ KE

Where

F_e = Coulomb force

r = Distance of the alpha particle from the gold foil

we get from F = k(Q₁·Q₂)/r²

W_{net} = \int\limits^{r_{min}}_{inf} {F_e} \, dr =(-1)× \int\limits^{r_{min}}_{inf} {\frac{k(2e)(79e)}{r^2} } \, dr

= -158·k·e² × \int\limits^{r_{min}}_{inf} {\frac{dr}{r^2} } \,⇒ -158·k·e²[\frac{1}{r} ]^{r_{min}} _{inf} = \frac{-158*k*e^2}{r_{min}} = -KE₁

or r_{min} = (158·k·e²)/KE₁

Where Δ KE = - KE₁ as KE₂ = 0, particle at rest

Solving gives

r_{min} =\frac{158*(8.988*10^9\frac{Nm^2}{C^2} )(1.602*10^{-19}C)^2}{6Mev*\frac{1.602*10^{-19}J}{eV} *1000000}     = 3.79 × 10⁻¹⁴ m

NISA [10]3 years ago
4 0

Answer:

r = 3.79 10⁻¹⁵ m

Explanation:

For this exercise we can use energy conservation

Starting point. Far from the fixed core

       Em₀ = K

Final point

        Em_{f} = U = k q1 q2 / r

           

      Em₀ =Em_{f}

       K = k q₁ q₂ / r

       r = k q₂ q₂ / K

Data

     K = 6 Mev = 6 106 eV (1.6 10⁻¹⁹ J / 1 eV) = 9.6 10⁻¹³ J

    Helium

    q₁ = 2 e = 2 1.6 10⁻¹⁹ C = 3.2 10⁻¹⁹ C

     Gold

    q2 = 79 e = 79 1.6 10⁻¹⁹ = 126.4 10⁻¹⁹ C

Let's calculate

      r = 8.99 10⁸ 3.2 10⁻¹⁹ 126.4 10⁻¹⁹ / 9.6 10⁻¹³

      r = 3.79 10⁻¹⁵ m

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Hitman42 [59]

For vertical motion, use the following kinematics equation:

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Given values:

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Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

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8 0
3 years ago
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Answer:

Option B. 6.25 J/S

Explanation:

Data obtained from the question include:

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Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

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magnetic field strength, B = 0.425 T

Induced peak emf = NABω

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Answer:

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h = P.E. / mg

h = 14000 / 40 × 9.8

h = 14000 / 392

h = 35.7

Therefore the canon ball was 35.7 meters  high.

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