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).
To determine the kinetic energy of an electron on a circular orbit around a nucleus, we can use the formula KE = (m*e^2)/(2*h^2*n^2), where m is the mass of the electron, e is the elementary charge, h is the Planck constant, and n is the principal quantum number (also equal to the number of the circular orbit). In this case, the principal quantum number is equal to 1 since we are dealing with the first orbit. Plugging in the values, we get KE = (9.11*10^-31*1.6*10^-19)/(2*6.626*10^-34*1^2) = 2.18*10^-18 J. To convert this to electron volts, we multiply by 6.24*10^18, giving us a final answer of 13.6 eV. This is the minimum amount of energy required to remove the electron from the atom, known as the ionization energy. In other words, this is the amount of energy that the electron possesses on this particular orbit. Since the question asks for the answer in multiples of 10^19, we have to multiply 13.6 by 10^19, giving us a final answer of 1.36*10^20. So the electron on this orbit has a kinetic energy of 1.36*10^20 Joules (13.6 eV). I hope this helps to understand the concept of kinetic energy in the hydrogen atom model!
