- #1

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[tex]

\lambda ' - \lambda = \frac{h}{m_{e} c} (1-cos \theta )

[/tex]

work even when [tex] \theta = 180^{\circ}[/tex]?

- Thread starter snoopies622
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- #1

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- 15

[tex]

\lambda ' - \lambda = \frac{h}{m_{e} c} (1-cos \theta )

[/tex]

work even when [tex] \theta = 180^{\circ}[/tex]?

- #2

malawi_glenn

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yeah sure, why shouldn't it?

- #3

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Maybe the mismatch is just due to rounding..

- #4

malawi_glenn

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Maximum transfer to electron is when E_prime is minimum

- #5

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At 180°, (E-E')/E = a/(1+a)

Maybe the mismatch is just due to rounding..

where:

E = photon's energy before scattering

E' = photon's energy after scattering

a = 2hv/mc^2 (v = photon's frequency before scattering).

The maximun value of (E-E')/E is 100% for: a = +oo, that is for an infinite energy of the photon before scattering.

What did you and your book get, instead?

Last edited:

- #6

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The book asks, "for the Compton effect, determine the percentage of energy transferred to the recoil electrons that acquire the maximum kinetic energy when [tex] \lambda = 1.21 x 10^{-10}m[/tex]."

I computed [tex]1-\frac {\lambda}{\lambda '}[/tex]

where [tex]\lambda ' = \lambda + \frac{h}{m_0 c}(1-cos \theta )[/tex]

and [tex]\theta = 180^{\circ}[/tex] since that maximizes [tex]1-cos \theta[/tex].

My answer: 3.85% (rounded from 3.85294..)

Book's answer: 3.93%.

I computed [tex]1-\frac {\lambda}{\lambda '}[/tex]

where [tex]\lambda ' = \lambda + \frac{h}{m_0 c}(1-cos \theta )[/tex]

and [tex]\theta = 180^{\circ}[/tex] since that maximizes [tex]1-cos \theta[/tex].

My answer: 3.85% (rounded from 3.85294..)

Book's answer: 3.93%.

Last edited:

- #7

jtbell

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- #8

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I guess that happens. Thanks for stepping in. I just looked at the Compton scattering section of another textbook and saw, "the scattered wavelength is angle-dependent and is greatest for scattering in the backward direction ([tex] \theta = 180^{\circ}[/tex])." So there you go.I suspect that whoever calculated the book's answer either used less-precise values for the constants, or rounded off one or more intermediate steps.

- #9

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lambda = 1.21*10^(-10) m

m =9.11*10^(-31) kg

c = 3.00*10^8 m/s

h = 6.63+10^(-34) J*s

I get 3.90% if I take m = 9.00*10^(-31) kg

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