High-speed electrons are used to probe the interior structure of the atomic
nucleus. For such electrons the expression \(\lambda = h/p\) still holds, but we
must use the relativistic expression for momentum, \(p = mv/\sqrt{1 -
v^2/c^2}\). (a) Show that the speed of an electron that has de Broglie
wavelength \(\lambda\) is $$v = \frac{c}{\sqrt1+(mc\lambda/h)^2} $$ (b) The
quantity \(h/mc\) equals 2.426 \(\times\) 10\(^{-12}\) m. (As we saw in Section
38.3, this same quantity appears in Eq. (38.7), the expression for Compton
scattering of photons by electrons.) If \(\lambda\) is small compared to \(h/mc\),
the denominator in the expression found in part (a) is close to unity and the
speed \(v\) is very close to c. In this case it is convenient to write \(v = (1 -
\Delta)c\) and express the speed of the electron in terms of rather than v.
Find an expression for \(\delta\) valid when \(\lambda \ll h mc\). [\(Hint:\) Use
the binomial expansion (1 + \(z)^n = 1 + nz + [n(n - 1)z^2/2] + \cdots\) g,
valid for the case 0 z 0 6 1.4 (c) How fast must an electron move for its de
Broglie wavelength to be 1.00 \(\times\) 10\(^{-15}\) m, comparable to the size of
a proton? Express your answer in the form \(v =(1 - \Delta)c\), and state the
value of \(\Delta\)