p39 078

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78.

(a) Using the result of problem 3,

E =

hc

λ

=

1240 nm

·eV

10.0

× 10

3

nm

= 124 keV .

(b) The kinetic energy gained by the electron is equal to the energy decrease of the photon:

E

=



hc

λ



= hc



1

λ

1

λ + ∆λ



=



hc

λ

 

λ

λ + ∆λ



=

E

1 +

λ

λ

=

E

1 +

λ

λ

C

(1

cos φ)

=

124 keV

1 +

10.0 pm

(2.43 pm)(1

cos 180

)

=

40.5 keV .

(c) It is impossible to “view” an atomic electron with such a high-energy photon, because with the

energy imparted to the electron the photon would have knocked the electron out of its orbit.


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