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Chapter 2: Problem 160

One of the emission spectral lines for \(\mathrm{Be}^{3+}\) has a wavelength of \(253.4 \mathrm{nm}\) for an electronic transition that begins in the state with \(n=5 .\) What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one- electron ions. Don't forget the \(Z\) factor: \(Z=\) nuclear charge \(=\) atomic number.

Short Answer

Expert verified
The principal quantum number of the lower-energy state corresponding to the given emission in the $\mathrm{Be}^{3+}$ ion is approximately \(2\).

Step by step solution

01

1. Recall the formula for the energy difference in the Bohr model.

We will use the formula for the energy difference between two energy levels in the Bohr model: \[ ΔE = -\dfrac{2\pi^2e^4Z^2}{h^2}\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right) \] where \(ΔE\) is the energy difference, \(e\) is the elementary charge, \(Z\) is the atomic number, \(h\) is the Planck constant, and \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final states, respectively. Since we know the initial state's principal quantum number and the ion's atomic number, we can rearrange the formula to solve for the final state's principal quantum number.
02

2. Convert the emission wavelength to energy difference.

We are given the emission wavelength \(λ = 253.4\,\text{nm}\) and need to convert it to the energy difference \(ΔE\). We can use the formula: \[ ΔE = \dfrac{hc}{λ} \] where \(c\) is the speed of light. Plugging in the values, we get: \[ ΔE = \dfrac{(6.626\times10^{-34}\,\text{J}\cdot\text{s})(3\times10^8\,\text{m/s})}{253.4\times10^{-9}\,\text{m}} = 7.845\times10^{-19}\,\text{J} \]
03

3. Substitute the variables and solve for the final state's principal quantum number.

Now, we will substitute the obtained value of \(ΔE\), the atomic number \(Z = 4\) (for beryllium), initial state's principal quantum number \(n_1 = 5\), and known constants into the energy difference formula, then solve for the final state's principal quantum number \(n_2\). Rearrange the formula for \(n_2\): \[ n_2^2 = \dfrac{1}{\frac{2\pi^2e^4Z^2}{h^2}\frac{1}{n_1^2}+\frac{2\pi^2e^4Z^2}{h^2ΔE}} = \dfrac{1}{\frac{1}{25}+\frac{2\pi^2(1.6\times10^{-19}\,\text{C})^4(4)^2}{(6.626\times10^{-34}\,\text{J}\cdot\text{s})^2(7.845\times10^{-19}\,\text{J})}} \] After solving the equation, we get: \[ n_2^2 \approx 4.002 \]
04

4. Find the principal quantum number of the lower-energy state.

Now, we can find the final state's principal quantum number by taking the square root of the value obtained above: \[ n_2 = \sqrt{4.002} \approx 2 \] Therefore, the principal quantum number of the lower-energy state corresponding to this emission is \(2\).

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Most popular questions from this chapter

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=4\) b. \(n=5, m_{\ell}=+1\) c. \(n=5, m_{s}=+\frac{1}{2}\) d. \(n=3, \ell=2\) e. \(n=2, \ell=1\)

It takes \(476 \mathrm{kJ}\) to remove 1 mole of electrons from the atoms at the surface of a solid metal. How much energy (in kJ) does it take to remove a single electron from an atom at the surface of this solid metal?

The successive ionization energies for an unknown element are \(I_{1}=896 \mathrm{kJ} / \mathrm{mol}\) \(\overline{I_{2}}=1752 \mathrm{kJ} / \mathrm{mol}\) \(I_{3}=14,807 \mathrm{kJ} / \mathrm{mol}\) \(I_{4}=17,948 \mathrm{kJ} / \mathrm{mol}\) To which family in the periodic table does the unknown element most likely belong?

Cesium was discovered in natural mineral waters in 1860 by R. W. Bunsen and G. R. Kirchhoff, using the spectroscope they invented in \(1859 .\) The name came from the Latin caesius ("sky blue") because of the prominent blue line observed for this element at \(455.5 \mathrm{nm} .\) Calculate the frequency and energy of a photon of this light.

Rank the elements Be, B, C, N, and O in order of increasing first ionization energy. Explain your reasoning.
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