We know from optics that when light of wavelength shines though a slit of width

ID: 1693049 • Letter: W

Question

We know from optics that when light of wavelength shines though a slit of width a, a diffraction pattern is produced, whose first minima appear at an angle given by a sin = lambda. If the first minima are seen at = pi/12. and the wavelength is = 550 nm, what is the width of the slit? If the diffraction pattern is observed at a distance L = 1.5 m away from the slit, how big is the diffraction pattern? Here, electrons are accelerated from rest by a potential V0 = 106V. What is the final momentum p of such electrons? Take mec2 = 0.511 MeV. The electrons in part (b) pass through a slit of width a = 100 pm. What is the width of the diffraction pattern for such electrons if the diffraction patten is observed at L = 1.5 m away from the slit?

Explanation / Answer

a) at ? = p / 12 and wave length ? = 550 nm width of the slit a = ? / sin ? = 550 *10^ -9 m / sin 15^ o = 2.125*10^ -6 m y = L sin ? = 1.5 * sin 15 = 0.388 m b) electrons are accelerated from rest by a potential V_o = 10^ 6 V kinetic energy of electrons k.E = q V _ o = ( 1.6*10^ -19 C ) ( 10^ 6 )V = 10^ 6 eV total energy E = 0.511*10^ 6 + 10^ 6 = 1.511 MeV final momentum og electrons P =E / c = 0.5*10^ -5 kg m /s c) wave length of incident electrons ? = h c / E = ( 6.625*10^ -34)( 3*10^ 8 m/s ) / ( 1.511 *10^ -13 J ) = 13.15*10-13 m y = ? L /a = ( 1.315 *10^ -12) ( 1.5 ) / 100 *10^ -12 ) = 0.019 m

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We know from optics that when light of wavelength shines though a slit of width

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