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In an early model of the hydrogen atom (the * Bohr model*), the electron orbits the proton in uniformly circular motion. The radius of the circle is restricted (

*quantized*) to certain values given by

*r* = *n*^{2}*a*_{0}, for *n* = 1, 2, 3, …,

where *a*_{0} = 52.92 pm. What is the speed of the electron if it orbits in (a) the smallest allowed orbit and (b) the second smallest orbit? (c) If the electron moves to larger orbits, does its speed increase, decrease, or stay the same?

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- This transmission electron microscope (TEM) image of coronavirus can be taken using a beam of electrons accelerated from rest through a potential difference of 25 kV. What is the final speed of the electrons? Provide the answer: . x 108 m/s
*arrow_forward*Orbital plane change: A satellite is launched due east from the Kennedy Space Center (latitude 28.6º). Following orbit insertion and a circularizing second burn with no plane change, the satellite is in a circular orbit at an altitude of 540 km. Its final destination is a circular orbit at 15,000 km with an inclination of 0 degrees. a. Using two Hohmann transfers, what total Δv is required to move the satellite from the initial orbit to the higher circular at the same inclination? Include a sketch showing the initial orbit, the transfer orbit and the final orbit. Include the location(s) of the burn(s). b. Calculate the Δv required to change the orbital inclination from its initial value to zero degrees in both the initial circular orbit and the higher circular orbit. Where would you want to make the change to the orbital inclination?*arrow_forward*where 1 eV = 1.602 × 10-19 J. Express the neutron’s kinetic energy in electron volts. b) In nuclear physics, it is convenient to express the energy of particles in electron volts (eV), 2) A neutron with a mass of 1.7 × 10-27 kg passes between two points in a detector 6 m apart in a time interval of 1.8 x 10-4 s. In the tendon at this pon a) Find the kinetic energy of the neutron in joules*arrow_forward* - Q 1: Let å = -2 ỉ +3 j +6 k and b = 4 i + 6 j , Find ( 2 å + 3 b ,1-ål, å * å , å * b_ )
*arrow_forward*Helpful information: (1) An alpha particle is a helium nucleus, (2) e = 1.6 × 10-¹⁹ C, (3) k₂ = 9.0 × 10⁹ Nm² C-2, (4) 1nm = 1 × 10-⁹ m 1-An alpha particle lies on the x-axis, a distance of 1.0 nanometer from a proton (in this set-up, the alpha particle is at the origin while the proton is in the positive direction). Which of the following choices below represents the magnitude of the electric force on the alpha particle? (a) 2.3 × 10-10 N (b) 4.6 × 10-10 N (c) 2.3 x 10-19 N (d) 4.6 x 10-19 N cing the voltage so following insta choices below at a time! 1.00 s?*arrow_forward*Suppose an electron (q = - e = - 1.6 x 10¬19 C,m=9.1 × 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K + U = 0 K = -U mv and using the formula for potential energy above, we arrive at an equation for speed: 2 Since K= v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=*arrow_forward* - Determine 1, 12, and 13, given that &, = 5 V, E2 = 10 V, E3 = 12 V, R1 = 2 0, R2 = 3 0, and R3 = 4 0. E, B E, = direction (left or right): direction (left or right): 13 = direction (left or right):
*arrow_forward*Consider an electron orbiting around a proton with an orbital radius of R=8.48⋅10−10R=8.48⋅10-10 m. What is the orbital frequency of the electron motion? Use me=9.11×10−31me=9.11×10-31 kg, e=1.6×10−19e=1.6×10-19 C, and k=9×109k=9×109 Nm2/C2. The frequency, f0 = Units . By how much would this frequency increase (assume the same orbital radius) if an external magnetic field of B = 0.5 T is applied to the system along the the electron axis of rotation? The increase in the frequency , Δf = Units .*arrow_forward*A proton is accelerated through a potential difference of 6 MV. (megavolts) from rest. Calc. the final velocity in m/s (a) 1.4 X 10' (b) 2.4 X 10' (c) 3.4 X 10' (d) 4.4 X 10'.*arrow_forward* - An electron is fired at a speed vi = 4.1 × 106 m/s and at an angle θi = 38.2° between two parallel conducting plates as shown in the figure. If s = 1.9 mm and the voltage difference between the plates is ΔV = 98.3 V, determine how close, w, the electron will get to the bottom plate. Put your answer in meters and include at 6 decimal places in your answer.
*arrow_forward*Suppose an electron (q= -e= -1.6 x 10-19 C,m=9.1x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K +U = 0 K = -U Since K- and using the formula for potential energy above, we arrive at an equation for speed: v = ( 51/2 Plugging in values, the value of the electron's speed is: V= x 107 m/s*arrow_forward*So Prof. P takes 2 handy metal plates and places them parallel to each other. He carefully adjusts them so they are 8.53 cm apart from each other and whips up a power supply to create a electric potential of 93 Volts between the plates. He notices there is a tiny hole on the positive plate so he decides to shoot a strangely handy electron straight through the hole at 3,950,851 m/s. Why? Who knows? How far in meters will the hapless electron travel from the positive plate until it stops?*arrow_forward*

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