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Calculate the orbital for the hydrogen atom that contains an electron with an energy of -4.45 × 10-²⁰ J.

User Schoola
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1 Answer

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Final answer:

The electron is in the 7th Bohr orbit of the hydrogen atom.

Step-by-step explanation:

The energy of an electron in a hydrogen atom is given by the formula E = -13.6 eV/n^2, where n is the principal quantum number or the number of the Bohr orbit.

In this case, we can rearrange the formula to solve for n^2. Given that the energy of the electron is -4.45 × 10^-20 J, we can convert it to electron volts (eV) using the conversion factor 1 eV = 1.6 × 10^-19 J.

Substituting the values into the formula, we have: -4.45 × 10^-20 J = -13.6 eV/n^2. Solving for n^2,

we find:

n^2 = (-13.6 eV)/(-4.45 × 10^-20 J).

Converting -4.45 × 10^-20 J to eV, we get -4.45 × 10^-20 J × (1 eV/1.6 × 10^-19 J) = -0.278 eV.

Substituting this value back into the formula, we have:

n^2 = (-13.6 eV)/(-0.278 eV) = 48.92.

Taking the square root of both sides, we find:

n = √(48.92) ≈ 6.99.

Therefore, the electron is in the 7th Bohr orbit.

User Trenton
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