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sukhopar [10]
3 years ago
7

Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the lead nucleus remains

stationary. Assume the alpha particles are initially very far from a stationary lead nucleus.
Chemistry
1 answer:
kiruha [24]3 years ago
7 0

Answer:

Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)

If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C

Using dc = (1/4πεo)qQ/Eα we have

dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm

Note: 1meter = 10^15fentometer

Explanation:

This is well inside the atom but some eight nuclear diameters from the centre of the lead nucleus.

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Luba_88 [7]

Answer:

Q = 19255.6 j

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of water = 46 g

change in temperature = ΔT = 100-0.0 = 100 °C

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Solution:

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Q = m.c. ΔT

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Q = 19255.6 j

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Explanation:

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