Home > 0877 > GR0877 Problem 85

GR0877 Problem 85

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) As one can check for himself/herself the constant A is equal to \displaystyle \sqrt{\frac{2}{L}} (this is from normalization condition \displaystyle  \int_0^L \psi^2(x) dx = 1). The probability in question is: \displaystyle \frac{2}{L} \int_{L/3}^{2L/3} \sin^2 \left(\frac{3 \pi x}{L}\right)dx = \frac{1}{L} \left[ \int_{L/3}^{2L/3} dx - \int_{L/3}^{2L/3} \cos \left( \frac{6 \pi x}{L} \right)dx \right] = \frac{1}{L} \left[ \frac{L}{3} -0 \right] = = \displaystyle \frac{1}{3}.

Found a typo? Comment!

Jump to the problem

1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Categories: 0877 Tags: , , , ,
  1. 14.10.2011 at 01:55

    A= \sqrt{2/L}, not just 2/L. Using 2/L results in the probability being equal to 2/3L, which is wrong. To get A, one needs to remember to square A, resulting in A^2=2/L, so therefore A=\sqrt{2/L}.

    • 14.10.2011 at 12:16

      Thank you, Oliver. This typo has been fixed.

  2. WobbleWheel
    07.11.2011 at 05:25

    Easier: if you recall that the wavefunctions for the infinite square well go as A\sin(n \pi x/L), then one can see that n=3. The probability distribution for the n=3 state is three humps. The given range, x=L/3 to x=2L/3, bounds the middle hump. Thus the probability is 1/3!

    • Steve
      11.11.2011 at 00:07

      That’s the way to do it in 1.7 minutes. I don’t think ETS expects you do that integral in the given time.

      • Kevin Tah
        20.10.2015 at 22:11

        I agree! There has got to be a way to be able to do this in 1.7 minutes. It seems a rather long problem to do it conventionally.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: