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.