In a Young’s double-slit experiment that uses electrons,the angle that locates t

Question

In a Young’s double-slit experiment that uses electrons,the angle that locates the first-order bright fringes isA = 8.1 x 10-4 degrees whenthe magnitude of the electron momentum is pA =1.2 x 10-22 kg*m/s. With the same double slit, whatmomentum magnitude pB is necessary so that anangle of B = 16.2 x 10-4degrees locates the first-order bright fringes?
I'm not sure where to start. In a Young’s double-slit experiment that uses electrons,the angle that locates the first-order bright fringes isA = 8.1 x 10-4 degrees whenthe magnitude of the electron momentum is pA =1.2 x 10-22 kg*m/s. With the same double slit, whatmomentum magnitude pB is necessary so that anangle of B = 16.2 x 10-4degrees locates the first-order bright fringes?
I'm not sure where to start.

Explanation / Answer

m = d sin 1 / 2 = sin1 / sin 2   m= 1 and d is the same for both cases = h / (m v) = h / p     the DeBroglie wavelength Then 1 / 2 =p2 / p1 = sin 1 /sin 2    p2 = p1 sin1 / sin 2    p2 = 1.2 * 10E-22 * sin 8.1 * 10E-4 / sin 16.2* 10E-4 Since is so small sin =   to a high degree of accuracy (No need to convert to radians because the ratio will be thesame) p2 = 1.2 * 10E-22 * 1 / 2 = 6 * 10E-23 kg-m /sec p2 = 1.2 * 10E-22 * sin 8.1 * 10E-4 / sin 16.2* 10E-4 Since is so small sin =   to a high degree of accuracy (No need to convert to radians because the ratio will be thesame) p2 = 1.2 * 10E-22 * 1 / 2 = 6 * 10E-23 kg-m /sec

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