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A parallel stream of monoenergetic electrons falls normally on a diaphragm with narrow square slit of width b=1.0 mu m. Find the velocity of the electrons if the width of the central diffraction maximum formed on a screen located at a distance l= 50cm from the slit is equal to Delta x= 0.36mm. |
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Answer» SOLUTION :The first MINIMUM in a Fraunhofer diffraction is given by (`B` is the width of the slit) `b sin theta= lambda` Here `sin theta=(Deltax//2)/(sqrtl^(2)+((Deltax)/(2))^(2))~~(Deltax)/(2l)` Thus `lambda=(bDeltax)/(2l)=(2pi ħ)/(mv)` so `V=(4pi ħl)/(mbDeltax)=2.02xx10^(6)m//s` |
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