What is the kinetic energy of an electron in the n = 2 Bohr orbit of the hydrogen atom?

Question

Quizplus · Accepted Answer

The kinetic energy of an electron in the n-th orbit of hydrogen atom is given by:
KE = -(13.6 eV)/n^2
where eV is electron volt and n is the principal quantum number.
For n = 2, KE = -(13.6 eV)/2^2 = -3.4 eV

Now, we need to convert this to joules as follows:
-3.4 eV x (1.602 x 10^-19 J/eV) = -5.45 x 10^-19 J (note the negative sign indicates that the electron is bound to the nucleus, so its kinetic energy is negative)

Therefore, the answer is C) 5.45 × 10^-19 J.

What Is the Kinetic Energy of an Electron in the N