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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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4 years ago
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The atomic number of manganese is 25; its mass number is 55.
Anuta_ua [19.1K]

Answer:

c) 30

Explanation:

In an atom , the relation between the subatomic particles of the atom are given as follows -

For a neutral atom ,

Atomic number = number of protons = number of electrons ,

Mass number = number of protons + number of neutrons,

Hence , from the given information of the question,

Atomic number of Manganese = 25 ,

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Now, to calculate the number of neutrons , the above relation can be used -

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Rearranging ,

number of neutrons = Mass number - number of protons

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number of neutrons = 55 - 25 = 30 .

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3 years ago
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worty [1.4K]
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Answer:

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Explanation:

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3 years ago
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DanielleElmas [232]

Answer:

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Explanation:

At higher temperatures, particles in an object moves faster.

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3 years ago
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