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Chapter 37: Problem 57

Green fluorescent protein (GFP) is a substance that was first ex- tracted from jellyfish; variants are used to "tag" biological molecules for study. The original "wild" GFP absorbs 395 -nm light, undergoing an upward transition to an excited state. Movement of a proton within the protein then excites it to 2.44 eV above the ground state. Photons emitted in the subsequent downward transition to the ground state provide a visual indication of the GFP's location as seen in a microscope. What's the wavelength of these photons?

Short Answer

Expert verified
After performing the calculation, we find that the wavelength of the emitted photons is approximately 508 nm.

Step by step solution

01

Conversion of Energy

Firstly, convert the energy from eV to Joules because the constant we are using (Planck's constant) is in joules. 1eV = \(1.602 \times 10^{-19}\) Joules. So, 2.44 eV = \(2.44 \times 1.602 \times 10^{-19}\) Joules.
02

Application of Formula

Now apply the formula E = h*c/λ. Here, E represents the energy of the photon, h represents Planck's constant, c represents the speed of light and λ represents the wavelength of the light. Rearrange the formula to find the wavelength. The formula becomes λ = h*c/E.
03

Substitution and Calculation

Next step involves substituting the values for the constants and the energy into the formula. h (Planck's constant) = \(6.626 \times 10^{-34}\) m²kg/s, c (speed of light) = \(3.00 \times 10^{8}\) m/s, E = \(2.44 \times 1.602 \times 10^{-19}\) Joules (from step 1). Substituting these values into the formula yields λ = \((6.626 \times 10^{-34}) \times (3.00 \times 10^{8}) / (2.44 \times 1.602 \times 10^{-19})\). Calculation will provide us with the correct answer.

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