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Chapter 27: Problem 31

What is the orbital radius of the electron in the \(n=3\) state of hydrogen?

Short Answer

Expert verified
Answer: The orbital radius of the electron in the n=3 state of hydrogen is approximately \(4.792 \cdot 10^{-10}\) meters.

Step by step solution

01

Recall Bohr's formula for the radius of an electron orbit in a hydrogen atom

The formula for the radius of an electron orbit in a hydrogen atom, according to Bohr's model, is given by: \(r_n = \frac{n^2·h^2·ε₀}{π·m_e·e^2}\) Where: - \(r_n\) is the radius of the orbit, which we want to calculate - \(n\) is the principal quantum number - \(h\) is Planck's constant (\(6.626 \cdot 10^{-34}\) J·s) - \(ε₀\) is the permittivity of free space (\(8.854 \cdot 10^{-12}\) C²/N·m²) - \(m_e\) is the mass of the electron (\(9.109 \cdot 10^{-31}\) kg) - \(e\) is the electron charge (\(1.602 \cdot 10^{-19}\) C) - \(π\) is Pi (approximately \(3.14159\))
02

Substitute the given value of n and constants into Bohr's formula

We are given the value of n, which is 3. Therefore, we can substitute n, along with the constants, into the formula to solve for the orbital radius: \(r_3 = \frac{3^2·(6.626 \cdot 10^{-34})^2·(8.854 \cdot 10^{-12})}{π·(9.109 \cdot 10^{-31})·(1.602 \cdot 10^{-19})^2}\)
03

Perform the calculations

Now, we will compute the expression on the right side of the equation: \(r_3 = \frac{9·(6.626 \cdot 10^{-34})^2·(8.854 \cdot 10^{-12})}{3.14159·(9.109 \cdot 10^{-31})·(1.602 \cdot 10^{-19})^2} \approx 4.792 \cdot 10^{-10}\) m
04

State the result

The orbital radius of the electron in the n=3 state of hydrogen is approximately \(4.792 \cdot 10^{-10}\) meters.

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Most popular questions from this chapter

Nuclei in a radium-226 radioactive source emit photons whose energy is $186 \mathrm{keV} .$ These photons are scattered by the electrons in a metal target; a detector measures the energy of the scattered photons as a function of the angle \(\theta\) through which they are scattered. Find the energy of the \(\gamma\) -rays scattered through \(\theta=90.0^{\circ}\) and \(180.0^{\circ} .\)
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