Consider a hydrogen atom in the ground state. What is the energy of its electron?
2 answers:
Hello
I found the radius of the electron's orbit
but I couldn't get the correct energy at ground state using the electric potential energy formulas
the radius I found matches the accepted value of Bohr's hydrogen atom radius
Edit:
I found the energy, my calculator wasn't calculating the small numbers correctly, so I kept on getting zero.
the energy at ground state is 4.36*10^-18 J
I can't reupload a new picture of the solution. so if it's possible, you might want to repost your question and message me to answer it again.
btw the potential energy is:
((e^4)*m)/(4(Elipson0)*h^2)= 4.36*10^-18 J
I found the radius of the electron's orbit
but I couldn't get the correct energy at ground state using the electric potential energy formulas
the radius I found matches the accepted value of Bohr's hydrogen atom radius
Edit:
I found the energy, my calculator wasn't calculating the small numbers correctly, so I kept on getting zero.
the energy at ground state is 4.36*10^-18 J
I can't reupload a new picture of the solution. so if it's possible, you might want to repost your question and message me to answer it again.
btw the potential energy is:
((e^4)*m)/(4(Elipson0)*h^2)= 4.36*10^-18 J
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Answer:
Molality = 1.428m
Explanation:
Molality, m, is an unit of concentration defined as the ratio between moles of solute and kg of solvent:
Molality: Moles solute / kg solvent.
In the problem, water is the solvent (Is the compound in the higher quantity) whereas NaNO₃ is the solute.
Moles of solute: 0.3355moles NaNO₃.
Kg solvent:
Thus, molality of the solution is:
Molality = 0.3355 moles NaNO₃ / 0.2350kg
<h3>Molality = 1.428m</h3>