Electrons with a speed of 2.4 Solution i solved for 1.2 mm. de Broglie wavelengt

Question

Explanation / Answer

i solved for 1.2 mm. de Broglie wavelength (?) associated with electron of (mass-m , v speed) ?e = h /p = h /me*ve >> p is momentum, e>electron --------------------------------- h = 6.6*10^-34 Js m (electron) = 9.1*10^-31 kg ?e = 6.6*10^-34 / 9.1*10^-31 * 2.3*10^6 meter ?e = 0.3153 *10^-9 meter --------------------------------------… fringe width ?x-e = D * ?e / d >>> D is distance of screen, d = slit separation ?x-e = D * h / me * v *d -----(1) --------------------- fringe width (spacing) when electron >> neutrons with same v ?x-n = D * h / mn * v *d -----(2) >>>> n stands for neutron , e>electron dividing ?x-n /?x-e = (me/mn) = 0.00091x10^(-27) /1.6749 x 10^(-27) = 5.433*10^-4 ?x-n = ?x-e * 5.433*10^-4 = 1.2 *10^-3 (meter) 5.433*10^-4 ?x-n = 0.652 *10^-6 (meter) = 0.652 micrometers ---------------------------- for different speeds and ?x-e = ?x-n = 1.2 mm 1.2 = D * h / me * ve *d 1.2 = D * h / mn * vn *d dividing 1 = (mn/me) (vn/ve) vn = ve* (me/mn) = 2.3*10^6 * 5.433*10^-4 = 1249.6 m/s speed of neutrons (vn)

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.