Question

In the Bohr model of the hydrogen atom, an electron (massm=9.1×10−31kg) orbits a proton at a distance of 5.3×10−11m. The proton pulls on the electron with an electric force of 8.2×10−8N. How many revolutions per second does the electron make?
Express your answer with the appropriate units.

Answer 1

Answer:

$\omega = 6.557 \times 10^{16}\ rev/s$

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)

      $F = m_e\omega^2 r$

      $\omega = \sqrt{\dfrac{F}{m_e\ r}}$

      $\omega = \sqrt{\dfrac{8.2 \times 10^{-8}}{9.1 \times 10^{-31}\times 5.3 \times 10^{-11}}}$

      $\omega = \sqrt{0.17 \times 10^{34}}$

       ω = 4.12 x 10¹⁶ rad/s

$\omega = \dfrac{4.12 \times 10^{16}}{2\pi}$

$\omega = 6.557 \times 10^{16}\ rev/s$