Answer:
[tex]\omega = 6.557 \times 10^{16}\ rev/s[/tex]
Explanation:
GIVEN,
mass of electron = 9.1 x 10 kg
Radius = 5.3 x 10 m
pulling force = 8.2 x 10 N
Required centripetal for (Fe) for circular motion will be provided with electrical force (F)
[tex]F = m_e\omega^2 r[/tex]
[tex]\omega = \sqrt{\dfrac{F}{m_e\ r}}[/tex]
[tex]\omega = \sqrt{\dfrac{8.2 \times 10^{-8}}{9.1 \times 10^{-31}\times 5.3 \times 10^{-11}}}[/tex]
[tex]\omega = \sqrt{0.17 \times 10^{34}}[/tex]
ω = 4.12 x 10¹⁶ rad/s
[tex]\omega = \dfrac{4.12 \times 10^{16}}{2\pi}[/tex]
[tex]\omega = 6.557 \times 10^{16}\ rev/s[/tex]
The electron makes 6.56 × 10^15 rev/s.
The electric force between the electrons and protons in an atom is exactly balanced by the centripetal force that keeps the electron moving along a circular path in the Bohr atom .
Felectric = F centripetal
Recall that centripetal force(F) = mv^2/r
v = rω
F= m(rω)^2/r
F =mr^2ω^2/r
F = mrω^2
ω = √F/mr
ω = √8.2×10^−8 N/9.1×10^−31 kg × 5.3×10^−11 m
ω = 4.123 ×10^16 rad/s
If 1 rev/s = 2π rad/s
x rev/s = 4.123 ×10^16 rad/s
x = 6.56 × 10^15 rev/s
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