Chapter 36: Problem 31

Show that the wavelength \(\lambda\) in \(\mathrm{nm}\) of a photon with energy \(E\) in eV is \(\lambda=1240 / E\)

Short Answer

Expert verified

\(\lambda=1240/E\) equation is derived using Planck's equation in appropriate units for energy in \(eV\) and wavelength in \(nm\).

Step by step solution

Introduction of General Energy-Wavelength Equation

The connection between energy and wavelength of a photon can be described using Planck's equation which is \(E = hc/\lambda\), where \(h\) is Planck's constant, \(c\) is the speed of light and \(\lambda\) is the wavelength of the photon.

Convert Units of Planck's Constant

In order to prove the given relation, units should be consistent. Hence convert the Planck's constant from its original units \(J.s\) to \(eV.s\), using the equivalence \(1eV = 1.602176565(35)×10^{-19} J\). Hence, the value of \(h\) becomes \(h = 6.62607004 × 10^{-34} Js / 1.602176565(35)×10^{-19}\) which equals approximately \(4.135667696 × 10^{-15} eV.s\).

Convert Speed of Light Unit

Next, convert speed of light units from \(m/s\) to \(nm/s\) for consistency with the wavelength's unit. Remember that \(1 m = 10^9 nm\), so \(c = 3.00 × 10^{8} m/s\) becomes \(c = 3.00 × 10^{17} nm/s\).

Substitute Values and Solve

Replace \(h\) and \(c\) with their newly calculated values in \(E=hc/\lambda\). Thus, becomes \(E= 4.135667696 × 10^{-15} eV.s * 3.00 × 10^{17} nm/s / \lambda\), which simplifies to \(1.240700981 × 10^{3}/ \lambda\). Therefore, \(\lambda= 1.240700981 × 10^{3}/E\), and rounding the value on the right gives \(\lambda= 1240/E\).

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Show that the wavelength \(\lambda\) in \(\mathrm{nm}\) of a photon with energy \(E\) in eV is \(\lambda=1240 / E\)

Short Answer

Step by step solution

Introduction of General Energy-Wavelength Equation

Convert Units of Planck's Constant

Convert Speed of Light Unit

Substitute Values and Solve

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