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Chapter 7: Problem 32

Calculate the frequency (Hz) and wavelength (nm) of the emitted photon when an electron drops from the \(n=4\) to the \(n=2\) level in a hydrogen atom.

Short Answer

Expert verified
The frequency of the emitted photon is \(6.15 * 10^{14} Hz\) and the wavelength is \(487 nm\).

Step by step solution

01

Calculate the energy difference

First, calculate the energy difference using the formula for energy levels in a hydrogen atom, \[ -13.6 eV \cdot \left(1 - \frac{1}{4}\right)\]. The energy of the higher level \(n=4\) is \(-13.6/4^2 = -0.85 eV\), for the lower level \(n=2\) it is \(-13.6/2^2 = -3.4 eV\). So, the energy difference is \(-0.85 eV - (-3.4 eV) = 2.55 eV\). To convert the energy from electron volts to Joules, which is a more common unit in such calculations, use the conversion \(1 eV = 1.6*10^{-19} J\). So, the energy difference is \(2.55 eV \cdot 1.6*10^{-19} J/eV = 4.08*10^{-19} J\).
02

Calculate frequency

The frequency can be found now using the equation \(E = h \cdot f\). Here, Planck's constant \(h = 6.63*10^{-34} Js\). Rearranging for \(f\), we get \(f = \frac{E}{h} = \frac{4.08*10^{-19} J}{6.63*10^{-34} Js} = 6.15 * 10^{14} Hz\).
03

Calculate the wavelength

Finally, to calculate the wavelength, use the wave equation \(c = f \cdot \lambda\), where \(c = 3 * 10^{8} ms^{-1}\), the speed of light. Rearranging for \(\lambda\), we get \(\lambda = \frac{c}{f} = \frac{3 * 10^{8} ms^{-1}}{6.15*10^{14} Hz} = 4.87 * 10^{-7} m\). To convert the result to nanometers, a common unit in this context, multiply by \(10^{9} nm/m\). So, \(\lambda = 4.87 * 10^{-7} m \cdot 10^{9} nm/m = 487 nm\).

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Indicate which of the following sets of quantum numbers in an atom are unacceptable and explain why: (a) \(\left(1,0, \frac{1}{2}, \frac{1}{2}\right)\) (b) \(\left(3,0,0,+\frac{1}{2}\right)\) (c) \(\left(2,2,1,+\frac{1}{2}\right)\) (d) \(\left(4,3,-2,+\frac{1}{2}\right)\), (e) \((3,2,1,1)\).

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