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प्रश्न
The basic force acting on the alpha particles using Coulomb's law is ______.
पर्याय
`"Ze"^2/(4pi∈_0"r"^2)`
`(2"Ze"^2)/(4pi"r"^2)`
`(2"Ze"^2)/(4pi∈_0"r"^2)`
`(2"Ze"^2)/(pi∈_0"r"^2)`
MCQ
रिकाम्या जागा भरा
उत्तर
The basic force acting on the alpha particles using Coulomb's law is `underline((2"Ze"^2)/(4pi∈_0"r"^2))`.
Explanation:
The gold nucleus has a charge of (Ze), where Z is the atomic number of the atom (Z=79), and alpha particles have two units of positive charge (2e).
The nucleus and the alpha particles are both tiny enough to be termed point masses. Also, because gold nuclei are 50 times heavier than alpha particles, they are expected to remain stationary during the experiment. As a result, according to Coulomb's law, the force is
F = `1/(4pi∈_0) ((2"e")("Ze"))/"r"^2`
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