Recent content by jcvince17

  1. J

    Calculate Mass Defect of Nitrogen

    Homework Statement For the common isotope of nitrogen, 14N, calculate the mass defect. Homework Equations \DeltaM = Zmp + Nmn - M The Attempt at a Solution 7(1.00727646688)+7(1.00866491560)-14.003074 \DeltaM = .108516177 Masteringphysics says this is wrong. says i may...
  2. J

    Consider a Hydrogen atom in the ground state

    now i solved part D and Part E. thanks
  3. J

    Consider a Hydrogen atom in the ground state

    i found a new equation in the book. K = 13.6 ev / n^2 U = -27.2 eV / n^2 E = K + U i knew E = 13.6. I finally got them all to work out without having to deal with the error's in the exponents.
  4. J

    Consider a Hydrogen atom in the ground state

    well i am trying to work out the units only to make sure i have the equation right. but i am not ending with eV E0 = C^2/N x m m = kg e = C h = J x s (or N x m x s) n = unitless i get: (1/m)(kg x C^2 / s) i cant convert or cancell anything else out. this problem is driving me insane
  5. J

    Consider a Hydrogen atom in the ground state

    i tried doing the exponents seperately: for the me^4 part of equation (1.67 x 10^-27) x (1.602 x 10^-19)^4 = [ 1.67 x 1.602^4 ] x 10^[-27- (19x4] = 10.99 x 10^-103 that looks very wrong. and when i try to divide/multiply it by something i get an error in the calc's. now what am i...
  6. J

    Consider a Hydrogen atom in the ground state

    what is the correct mass of Hydrogen. I am using 1.66x10^-27. But when i do so and multiply it by e^4 i get 0. which makes everything 0 and this is incorrect.
  7. J

    Consider a Hydrogen atom in the ground state

    I fixed my equations. now using the first equation to find K. I used the following values: E0 = 8.854x10-12 m = 1.66*10-27 (i think thats right for Hydrogen) e = 1.602x10-19 h = 6.626x10-34 n = 1 Kn = (1/E02)(me4/8n2h2) plug in the values and i get K = 1.28x1022 x 12.17 K =...
  8. J

    Consider a Hydrogen atom in the ground state

    Homework Statement Consider a Hydrogen atom in the ground state Find: A- The Kinetic Energy, in eV B - The Potential Energy, in eV C - The Total Energy, in eV D - minimum energy required to remove the electron completely from the atom, in eV E - What wavelength does a photon with the...
  9. J

    Two Lens System - Image

    This is due at 11pm tonight, if anyone can please help me. I dont know where I am going wrong here. Thanks.
  10. J

    Two Lens System - Image

    tried it again lens 1: 1/f = 1/s' + 1/s 1/10 = 1/s' + 1/30 s' = 7.5 = @ x = -12.5 lens 2: 1/f = 1/s' + 1/s 1/8 = 1/s' + 1/32.5 s' = 10.6 = final image is at x = + 30.6 ???
  11. J

    Two Lens System - Image

    well, I have gotten several wrong answers and have only two chances left. edit - still working on it i found something about solving for a resultant focal point. F = 1 / f1 + 1 / f2 - d / f1f2 F = (1/10 + 1/8) - (40/10*8) F = -.275 if i then take this into my original equation...
  12. J

    Snell's Law Problem

    no problem...i am the same way sometimes. sometimes you just need a clear head to straighten it out.
  13. J

    Snell's Law Problem

    i do it i get: 1sin90 = 1.33sin(theta 2) 1 = 1.33sin(theta 2) 1/1.33 = sin(theta 2) .7518796992 = sin(theta 2) sin-1(.7518796992) = 48.75 degrees
  14. J

    Two Lens System - Image

    Homework Statement A compound lens system consists of two converging lenses, one at x= -20cm with focal length f1 = +10cm, and the other at x= +20cm with focal length f2 = +8cm . An object 1 centimeter tall is placed at x = -50cm. What is the location of the final image produced by the...
  15. J

    Image Size In A Mirror

    doing so i get: 1/-15 = 1/s' + 1/3.4 s' (head) = -2.77 1/-15 = 1/s' + 1/5 s' (feet) = -3.75 so i subtract these two numbers and have an answer of 0.98 m Thank You!!!
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