What is the energy of a hydrogen atom with a 1.70 nm diameter?

To find the energy of a hydrogen atom with a 1.70 nm diameter, we need to use the equation for the energy of an atom. The energy of an atom is given by the formula:

E = -(13.6 eV) / n^2

where E is the energy in electron volts (eV) and n is the principal quantum number.

To find the principal quantum number (n) for a hydrogen atom with a given diameter, we can use the formula:

n = (2r/a_0) - 1

where r is the radius of the atom and a_0 is the Bohr radius, which is approximately 0.0529 nm for hydrogen.

Given that the diameter of the hydrogen atom is 1.70 nm, the radius (r) is half the diameter, so r = 0.85 nm. Substituting this value into the formula, we have:

n = (2 * 0.85) / 0.0529 - 1
= 32.114 - 1
= 31.114

Now that we have the principal quantum number (n), we can plug it into the energy formula to find the energy (E):

E = -(13.6 eV) / (31.114^2)
≈ -0.013 eV

Therefore, the energy of a hydrogen atom with a 1.70 nm diameter is approximately -0.013 electron volts.