В ядерной модели атома водорода электрон вращается вокруг ядра (протон) по круговой орбите, радиус которой R = 5,3 • 10-11 м. Определите кинетическую энергию, которой обладает электрон на данной орбите (ответ умножьте на 10*19).
The kinetic energy of the electron in the hydrogen atom can be calculated using the formula KE = (1/2) * m * v^2, where m is the mass of the electron and v is its velocity. Since the electron is moving in a circular orbit, its velocity can be calculated using the formula v = (2 * pi * r) / T, where r is the radius of the orbit and T is the orbital period. Since the electron is in a stable orbit, it follows that the centripetal force acting on it (provided by the electrostatic force between the electron and the proton) is equal to the centrifugal force (provided by the motion of the electron). This can be expressed in the equation F = mv^2 / r = k * q^2 / r^2, where k is the Coulomb constant, q is the charge of the electron, and r is the radius of the orbit. Substituting the value of v calculated earlier and solving for v, we get v = (k * q / r)^1/2. Substituting this value of v in the formula for kinetic energy, we get KE = (1/2) * m * ((k * q / r)^1/2)^2 = k^2 * q^2 / (2 * m * r). Plugging in the known values for these variables (m = 9.11 * 10^-31 kg, r = 5.3 * 10^-11 m, q = 1.6 * 10^-19 C, k = 8.99 * 10^9 N*m^2/C^2) and multiplying by 10^19 to get the result in joules, we get the final answer: 27.2 * 10^19 J. This is the kinetic energy of the electron on the given orbit in the hydrogen atom.
