Home > 0877 > GR0877 Problem 97

## GR0877 Problem 97

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) According to Compton scattering formula, the difference in wavelength of the scattered $\lambda'$ and incident $\lambda$ photon is $\displaystyle \lambda' - \lambda = \frac{h}{mc}(1-\cos{\theta})$, where $\theta$ is the scattering angle. In our case, $\theta = 90^{\circ}$ and $\displaystyle \lambda' - \lambda = \frac{h}{mc}$. But $\displaystyle E = \frac{hc}{\lambda}$, so $\displaystyle E' = \frac{hc}{\lambda'}$, where $E'$ is the energy of the scattered photon. Thus, $\displaystyle \frac{hc}{E'} - \frac{hc}{E} = \frac{h}{mc}$, from which $\displaystyle E' = \frac{E \cdot mc^2}{E + mc^2}$.

Found a typo? Comment!

 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. 15.10.2011 at 13:12

2 small typos: you wrote that $E=hc/\nu$ but should be $E=hc/\lambda$.
Another typo is in the compton scattering formula: should be $h/mc$ rather than $\hbar/mc$

• 15.10.2011 at 13:57

Thanks! Both typos have been fixed.

2. 24.09.2014 at 21:15

For those that don’t remember (or know anything about) the Compton formula, you can solve with conservation laws. Momentum of the electron will be equal to $$E^2+E_F^2$$ (conservation of momentum). Plug into $$E+m=E_f+\sqrt{m^2+p_e^2}$$ to get the final answer.

3. 20.10.2014 at 12:43

Surely the easiasiasiest way is consider and extreme panorama where
$\lim_{m \to +\infty} E’ = E$
cuz the photon shouldn’t lose its energy due to the momentum conservation (it’s an elastic scattering).
E is the only answer that fits this prediction.

• 20.10.2014 at 12:49

OH…IT’S CALLED thomson scattering!
$\frac{h \cdot \nu}{m \cdot c^2} \ll 1$

4. 20.10.2014 at 12:45

Surely the easiasiasiest way is consider and extreme panorama where
$\lim_{m \to +\infty} E’ = E$
cuz the photon shouldn’t lose its energy due to the momentum conservation (it’s an elastic scattering).
E is the only answer that fits this prediction.