Question

The Bohr model of the hydrogen atom pictures the electron in its ground state orbiting the proton at a distance of 5.0×10⁻¹¹ m at a speed of 2.5×10⁶ m/s. What is the escape velocity in this model?

Answer 1

In the Bohr model of the hydrogen atom, the escape velocity can be calculated by equating the kinetic energy of the electron to the potential energy, resulting in an escape velocity of approximately 2.18 × 10⁶ m/s.

Step-by-step explanation:

In the Bohr model of the hydrogen atom, the escape velocity can be calculated by equating the kinetic energy of the electron to the potential energy at the outermost orbit. The kinetic energy is given by ½mv² and the potential energy is given by ke²/r, where m is the mass of the electron, v is the speed of the electron, k is Coulomb's constant, and r is the distance from the proton.

By setting these two energies equal to each other and solving for v, we can find the escape velocity. Let's plug in the values given in the question:

Radius of the orbit (r) = 5.0×10⁻¹¹ m
Speed of the electron (v) = 2.5×10⁶ m/s
Mass of the electron (m) = 9.11 × 10⁻³¹ kg

Using these values, we can calculate the escape velocity as:

v = √((2k/m) * (1/r))

Plugging in the given values and solving for v, we get:

v = √((2 * 8.99 × 10⁹ Nm²/C²) / (9.11 × 10⁻³¹ kg)) * (1 / (5.0×10⁻¹¹ m))

After evaluating this expression, the escape velocity comes out to be approximately 2.18 × 10⁶ m/s. Therefore, in the Bohr model, the escape velocity is approximately 2.18 × 10⁶ m/s.