Question

A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary, and the electron has a speed of 6.1 105 m/s. Find the radius between the stationary proton and the electron orbit within the hydrogen atom.

Answer 1

Answer:

Radius between electron and proton$= 6.804\times 10^{-10}m$

Explanation:

The motion of the electron is carried out in the orbit due to the balancing of the electrostatic force between the proton and the electron and the centripetal force acting on the electron.

The electrostatic force is given as = $\frac{kq_1q_2}{r^2}$

Where,

k = coulomb's law constant (9×10⁹ N-m²/C²)

q₁ and q₂  = charges = 1.6 × 10⁻¹⁹ C

r = radius between the proton and the electron

Also,

Centripetal force on the moving electron is given as:

=$\frac{m_eV^2}{r}$

where,

$m_e$ = mass of the electron (9.1 ×10⁻³¹ kg)

V = velocity of the moving electron (given: 6.1 ×10⁵ m/s)

Now equating both the formulas, we have

$\frac{kq_1q_2}{r^2}$ = $\frac{m_eV^2}{r}$

⇒$r = \frac{kq_1q_2}{m_eV^2}$

substituting the values in the above equation we get,

$r = \frac{9\times 10^{9}\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times (6.1\times 10^5)^2}$

⇒$r = 6.804\times 10^{-10}m$