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In fluorescence microscopy, an important tool in biology, a laser beam is absorbed by target molecules in a sample. These molecules are then imaged by a microscope as they emit longer-wavelength photons in quantum jumps back to lower energy levels, a process known as fluorescence. A variation on this technique is two-photon excitation. If two photons are absorbed simultaneously, their energies add. Consequently, a molecule that is normally excited by a photon of energy$E_{photon}$can be excited by the simultaneous absorption of two photons having half as much energy. For this process to be useful, the sample must be irradiated at the very high intensity of at least ${10}^{32}\raisebox{1ex}{$photons$}\!\left/ \!\raisebox{-1ex}{$m^{2}s$}\right.$ . This is achieved by concentrating the laser power into very short pulses ( $100fs$pulse length) and then focusing the laser beam to a small spot. The laser is fired at the rate of ${10}^8$pulses each second. Suppose a biologist wants to use two-photon excitation to excite a molecule that in normal fluorescence microscopy would be excited by a laser with a wavelength of $420nm$. If she focuses the laser beam to a$2.0-mm$-diameter spot, what minimum energy must each pulse have?

Step by step solution

01

Given information

We are given with radius $r=1.0×{10}^{-6}m$,$\mathrm{λ}=420nm$wavelength , $I={10}^{32}p/m^{2}s$intensity, pulse length$\mathrm{τ}=100×{10}^{-15}s$.

We need to find out minimum energy that each pulse have .

02

Simplify

To find minimum energy value , we will use the relation $E_{2y}=\frac{1}{2}{E}_{1y}$.

We will use the formula of energy , $E_{2y}=\frac{1}{2}\left(\frac{hc}{\mathrm{λ}}\right)$. Here$\mathrm{λ}$is wavelength

We will substitute the values, as we know the wavelength$E_{2y}=\frac{1}{2}\left(\frac{1240eVnm}{420nm}\right)=\raisebox{1ex}{$1.48eV$}\!\left/ \!\raisebox{-1ex}{$p$}\right.$

We will find energy pulse using the relation ,$E_pâ‰\yen I\mathrm{Ï„}{\mathrm{Ï€°ù}}^2{\mathrm{E}}_{2\mathrm{y}}$here $E_p$is energy pulse,$I$is intensity,

$\mathrm{Ï„}$is pulse length , $r$is radius .

Substituting all the values we get, $E_pâ‰\yen \left({10}^{32}p/m^{2}s\right)\left(100×{10}^{-15}s\right)\mathrm{Ï€}\left(\frac{1.6×{10}^{-19}\mathrm{J}}{1.0\mathrm{eV}}\right)â‰\yen 7.3\mathrm{pJ}$

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