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Andru [333]
3 years ago
12

You will find in Chapter 39 that the electrons cannot move indefinite orbits within atoms, like the planets in our solar system.

To see why, let us try to "observe" such an orbiting electron byusing a light microscope to measure the electron's presumed orbitalposition with a precision of, say, 8.2 pm(a typical atom has a radius of about 100 pm). The wavelength ofthe light used in the microscope must then be about 8.2 pm.
(a) What would be the photon energy of thislight?
1 keV

(b) How much energy would such a photon impart to an electron in ahead-on collision?
2 keV

(c) What do these results tell you about the possibility of"viewing" an atomic electron at two or more points along itspresumed orbital path? (Hint: The outer electrons of atoms arebound to the atom by energies of only a few electron-volts.)
Physics
1 answer:
Amanda [17]3 years ago
5 0

Answer:

1.. 151keV

2. 56.9keV

3. It isn't possible to view an electron with such a photon because of the high energy

Explanation:

See attached file for calculation

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The windowpanes are___________ a. opaque b. transparent c. absorbent .
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Answer:

The windowpanes are- transparent.

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Explanation:

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A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni
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Answer:

Explanation:

Theorem of Binomial Distribution will apply here.

n = 29 , p = .67 , q = 0.33

mean = np = 29 x .67 = 19.43

Standard Deviation = √npq

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3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

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