Question

Imagine you are a world-famous scientist attempting to label chromosomes in order to study the detailed mechanics of recombination during meiosis I. In order to achieve the maximal possible resolving power, you would want to use a fluorophore with excitation wavelength of:

		719 nm (red)
		602 nm (yellowish-orange)
		532 nm (green)
		1 mm (infrared)

Answer 1

Resolving power is expressed by the formula 1.22 λ/D, where λ is the wavelength of light and Dis the diameter pof the aperture.Here it is considered that the D is same. the lesser the wavelength the greater is the resolving power.

if the excitation wavelength is less it emits at a higher wavelength.

The least excitation wavelength in this question is 532nm(green). the resolving power is inversely proportional to the wavelength of the light used. So the light with least wavelength gives highest resolution.

In this case it is gree, 532nm(option C)