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
To determine the kinetic energy of an electron on a circular orbit in the nuclear model of a hydrogen atom, we need to use the formula KE = (1/2)mv^2, where m is the mass of the electron and v is its velocity. Since the electron is moving in a circular path, we can use the formula for the centripetal force F = mω^2R, where R is the radius of the orbit and ω is the angular velocity. Equating the centrifugal force with the electrostatic force between the electron and the proton, we can get the expression for the angular velocity ω = v/R. Substituting this into the formula for KE, we get KE = (1/2)mv^2 = (1/2)(mωR)^2 = (1/2)m(v^2/R^2)R^2 = (1/2)mv^2R^2 = mω^2R^2/2. Now we can substitute the value of R = 5.3•10^-11 m and the mass of an electron m = 9.11•10^-31 kg into the formula to get KE = (9.11•10^-31 kg)(3.29•10^15 s^-1)^2(5.3•10^-11 m)^2/2 = 2.18•10^-18 J. Therefore, the kinetic energy of an electron on a circular orbit in the nuclear model of a hydrogen atom is 2.18•10^-18 J. Don't worry, even though it looks like a small number, it is actually a very significant value in the atomic world!
