Question

In the Bohr model of the atom, an electron can be thought of a small sphere that rotates around the nucleus. In a hydrogen atom, an electron (me=9.11 x 10^-31 kg) orbits a proton at a distance of 5.3 x 10^-11 m from the proton. If the proton pulls on the electron with a force of 9.2 x 10^8 N, how many revolutions per second does the electron make?

6. In the Bohr model of the atom, an electron can be thought of a small sphere that rotates around the nucleus. In a hydrogen atom, an electron (me-9.11 × 10-31 kg) orbits a proton at around the nucleus. In a hydrogen atom, an electron (me 9.11 x 10-31 kg) orbits a proton at distance of 5.3 × 10-11 rn from the proton. If the proton pulls on the el 9.2 x 108 N, how many revolutions per second does the electron make? ectron with a force of

Answer 1

m_e = 9.11 times 10^-31 k g d = 5.3 times 10^-11 m F = 9.2 times 10^8 N This problem could be treated like a circular motion, the attracting force due to the electromagnetism is the centripetal force. The centripetal force is related to the angular velocity through the formula F_c = m middot w^2 middot R as: m = mass w = angular velocity R = radius We need to now the angular velocity (how many revolutions per second does the electron make) clearing w w^2 = F_c/m middot R w = squareroot F_c/m middot R w = squareroot 9.2 times 10^8 N/9.11 times 10^-31 k g middot 5.3 times 10^-11 m w = 4.365 times 10^24 rev/s

$d=5.3\times 10^{-11}m$

$F=9.2\times 10^8 N$

This problem could be treated like a circular motion, the attracting force due to the electromagnetism is the centripetal force.

The centripetal force is related to the angular velocity through the formula

$F_c=m.\omega ^2.R$

as:

m= mass

w= angular velocity

R= radius

We need to now the angular velocity (how many revolutions per second does the electron make)

clearing w

$\omega ^2=\frac{F_c}{m.R}$

$\omega =\sqrt{\frac{F_c}{m.R}}$

$\omega =\sqrt{\frac{9.2\times10^8N}{9.11\times10^{-31}Kg.5.3\times10^{-11}m}}$

$\omega =4.365\times10^{24}rev/s$

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