Search results

  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!!!
  16. J

    Image Size In A Mirror

    Homework Statement A guy named Joe, who is 1.6 meters tall, enters a room in which someone has placed a large convex mirror with radius of curvature equal to 30 meters. The mirror has been cut in half, so that the axis of the mirror is at ground level. As Joe enters the room, he is 5...
  17. J

    Wavelength, Frequency, & Speed of Light in Different Media

    thanks you. i will begin working out the substitution.
  18. J

    Wavelength, Frequency, & Speed of Light in Different Media

    Homework Statement A beam of light from a monochromatic laser shines into a piece of glass. The glass has thickness L and index of refraction n=1.5. The wavelength of the laser light in vacuum is L/10 and its frequency is f. In this problem, neither the constant c nor its numerical value...
  19. J

    Converting to nanometers

    yes that got the first problem correct. thanks
  20. J

    Converting to nanometers

    Homework Statement i just need to convert 2.8*1014 into nano meters Homework Equations \lambda = v / f The Attempt at a Solution \lambda = v / f \lambda = 2.0*108 / 700*10-9 (nanometers) \lambda = 2.8*1014 i am backwards on my conversion, i am horrible at it...
  21. J

    Simple Index of Refraction Question - chart

    from my book: " the index of refraction of an optical material, denoted as n, is the ration of the speed of light in a vacuum (c) to the speed of light in the material (v) n = c / v " ok so...light always travels more slowly in material....therefore it is always 1+ thanks for...
  22. J

    Simple Index of Refraction Question - chart

    Homework Statement what is the minimum value that the index of refraction can have? Homework Equations n1sin\Thetai = n2sin\Thetar The Attempt at a Solution my choices are: A - between 0 and 1 B - +1 C - -1 D - 0 all listed values in my given chart are...
  23. J

    Induced current through a resistor - with variable resistor and two coils

    my reasoning sound correct also? thanks for all your help! with me knowing the directions on coil 1 it made it easier to explain what I needed coil 2 to have
  24. J

    Induced current through a resistor - with variable resistor and two coils

    i filled in above. here is what was I got from class tonight. we have a current to the right and B to the left on coil 1. with the variable resistor it is decreasing I which in turn is decreasing B. If we have to "oppose" a decreasing B then the B of coil 2 needs to be to the left as...
  25. J

    Induced current through a resistor - with variable resistor and two coils

    if resistance is increased then the current (I) is decreased correct? I= V/R if the current is decreased in coil 1 then the B of coil 2 is decreased as well right? decreasing B filed to the left.
  26. J

    Induced current through a resistor - with variable resistor and two coils

    honestly i do not know. i was going on the flow of the current in coil (left) and assuming it went from a to b. i do not know anything about the Rvar. What does it do? i am not 100% sure how the current in one and two flows to the right. i just looked at the coils and saw its direction...
  27. J

    Induced current through a resistor - with variable resistor and two coils

    if this is true (what i posted above here) then would the current in coil loop two flow from b to a?
  28. J

    Induced current through a resistor - with variable resistor and two coils

    current to the right and magnetic field going to the left. right?
  29. J

    Induced current through a resistor - with variable resistor and two coils

    The current will flow from left to right correct? and magnetic field into the page? this new right hand rule has me off. I am to place my thumb in direction of B and curl my fingers into the induced I? therefore the current on coil 1 will flow in a clockwise direction?
Top