Question

che105(mwf) - worksheet #21 - photoelectric effect

1. when a compound containing cesium ions is heated in a bunsen burner flame, photons with an energy of $4.30 \times 10^{-19} \text{ j}$ are emitted. what color is the cesium flame?

Explanation:

Step1: Use the energy - wavelength formula

The energy of a photon is given by \(E = h
u=\frac{hc}{\lambda}\), where \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant), \(c=3\times 10^{8}\space m/s\) (speed of light), and \(\lambda\) is the wavelength. Rearranging for \(\lambda\), we get \(\lambda=\frac{hc}{E}\).
Substitute \(h = 6.626\times 10^{-34}\space J\cdot s\), \(c = 3\times 10^{8}\space m/s\), and \(E=4.30\times 10^{-19}\space J\) into the formula:
\(\lambda=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{4.30\times 10^{-19}}\)

Step2: Calculate the wavelength

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Convert meters to nanometers (\(1\space m = 10^{9}\space nm\)), so \(\lambda=462\space nm\)

Answer:

The wavelength of the emitted photons is \(462\space nm\), which corresponds to the blue color.