An electron is accelerated in a straight line through an electrical potential di

Question

An electron is accelerated in a straight line through an electrical potential difference produced by two metal plates, each with a small hole (so the electron can make it through them). Assume the electron is at rest when it starts at the left plate. The electron has a velocity of 5 times 10^-5[m/s] when it passes the right plate. Assume the potential of the right plate is 0[V]. Fill out the table. M_c = 9.1 times 10^-31 [kg] KE of electron after it passes through the right plate:______Voltage of the Right plate____0[V] given Voltage of the Left plate_____After the electron leaves the right plate it enters a region where a uniform magnetic field, perpendicular to the paper, is suddenly turned on. The electron makes a circular path with its angular velocity, omega, pointing out of the page (this means counter-clockwise motion when viewed on this page) The radius of the circular motion is 0.1[nm]. Fill out the table. Use symbols like to leftarrow downarrow uparrow rightarrow otime odot indicate directions. Direction of B-field: Magnitude of B-field:_____Magnitude of omega:______Direction of OMEGA:_odot_____Period of revolution: T =_______The magnetic field is eventually shut off and the electron slams into a stationary proton, making a hydrogen atom. Model the collision as A maximally inelastic collision. and calculate the final speed of the hydrogen atom. Ignore any rotation of the hydrogen atom. m_P = 1.7 times 10^-27 [kg]

Explanation / Answer

12 a) qV = 0.5mv^2
V = 0.7109 V
V left = 0.7109 V
b) From left hand rule, B is out of the page
qvB = mv^2/r
B = 0.0239 T
c) m1u1 + m2u2 = Mv
u1 = 0
so, v = 267.50

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