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標題:
physics
發問:
1.The neutrons in a parallel beam, each having kinetic energy 1/40 eV, are directed through two slits 0.48 mm apart. How far apart will the interference peaks be on a screen 1.6 m away? [Hint: first find the wavelength of the neutron.]2.If an electron's position can be measured to an accuracy of 2.4 10-8... 顯示更多 1.The neutrons in a parallel beam, each having kinetic energy 1/40 eV, are directed through two slits 0.48 mm apart. How far apart will the interference peaks be on a screen 1.6 m away? [Hint: first find the wavelength of the neutron.] 2.If an electron's position can be measured to an accuracy of 2.4 10-8 m, how accurately can its speed be known?
最佳解答:
(1) First of all, we have to find out the velocity of each of the moving neutrons as follows: Consider that its k.e. is given by: (1/2)mv2 = (1/40) × 1.6 × 10-19 where m is the neutron mass = 1.67 × 10-27 kg We have: v = 2190 m/s Then we have to find out the wavelength of the particle wave associated with one neutron: λ = h/mv where λ and h are wavelength and Planck's constant respectively. Substituting the value, we have: λ = 6.63 × 10-34/(2190 × 1.67 × 10-27) λ = 1.81 × 10-10 m Finally, we apply the equation of Young's double slits experiment: s = λD/a where s = peaks separation on screen λ = wavelength D = slits-screen separation a = slits separation We have the separation between successive interference peaks as: s = (1.6 × 1.81 × 10-10)/(0.48 × 10-3) = 6.04 × 10-7 m (2) Using the Heisenberg's Uncertainty Principle: h = (△x)(△p) where h = Planck's constant △x = Uncertainty in position △p = Uncertainty in momentum We have: △p = (6.63 × 10-34)/(2.4 × 10-8) = 2.761 × 10-26 kg m/s So we have: △v = (2.761 × 10-26)/(9.1 × 10-31) where 9.1 × 10-31 kg is the mass of an electron. = 30.3 × 103 m/s
其他解答:
physics
發問:
1.The neutrons in a parallel beam, each having kinetic energy 1/40 eV, are directed through two slits 0.48 mm apart. How far apart will the interference peaks be on a screen 1.6 m away? [Hint: first find the wavelength of the neutron.]2.If an electron's position can be measured to an accuracy of 2.4 10-8... 顯示更多 1.The neutrons in a parallel beam, each having kinetic energy 1/40 eV, are directed through two slits 0.48 mm apart. How far apart will the interference peaks be on a screen 1.6 m away? [Hint: first find the wavelength of the neutron.] 2.If an electron's position can be measured to an accuracy of 2.4 10-8 m, how accurately can its speed be known?
最佳解答:
(1) First of all, we have to find out the velocity of each of the moving neutrons as follows: Consider that its k.e. is given by: (1/2)mv2 = (1/40) × 1.6 × 10-19 where m is the neutron mass = 1.67 × 10-27 kg We have: v = 2190 m/s Then we have to find out the wavelength of the particle wave associated with one neutron: λ = h/mv where λ and h are wavelength and Planck's constant respectively. Substituting the value, we have: λ = 6.63 × 10-34/(2190 × 1.67 × 10-27) λ = 1.81 × 10-10 m Finally, we apply the equation of Young's double slits experiment: s = λD/a where s = peaks separation on screen λ = wavelength D = slits-screen separation a = slits separation We have the separation between successive interference peaks as: s = (1.6 × 1.81 × 10-10)/(0.48 × 10-3) = 6.04 × 10-7 m (2) Using the Heisenberg's Uncertainty Principle: h = (△x)(△p) where h = Planck's constant △x = Uncertainty in position △p = Uncertainty in momentum We have: △p = (6.63 × 10-34)/(2.4 × 10-8) = 2.761 × 10-26 kg m/s So we have: △v = (2.761 × 10-26)/(9.1 × 10-31) where 9.1 × 10-31 kg is the mass of an electron. = 30.3 × 103 m/s
其他解答:
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