Question 38: Dual Nature of Matter and Radiation

Question

A proton is fired from very far away towards a nucleus with charge Q = 120e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is ____________. (Take the proton mass, $${m_p} = (5 	imes 3) 	imes {10^{ - 27}}$$ kg; $$h/e = 4.2 	imes {10^{ - 15}}$$ J.s/C; $${1 \over {4\pi {\varepsilon _0}}} = 9 	imes {10^9}$$ m/F; 1 fm = 1015 m.)

Accepted Answer

Let initial and final kinetic energies of the proton be K
i
and K
f
and corresponding potential energies be U
i
and U
f
. When proton is far away from the nucleus (r $$	o$$ $$\infty$$), its potential energy is
$${U_i} = \mathop {\lim }\limits_{r 	o \infty } {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over r} = 0$$.
At closest distance, the proton comes to rest momentarily, giving K
f
= 0. The potential energy at closest distance is
$${U_f} = {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over a}$$,
where a is the distance of closest approach. Since electrostatic force is conservative, total energy is conserved i.e., $${K_i} + {U_i} = {K_f} + {U_f}$$. Substitute the values to get
$${K_i} = {U_f} = {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over a}$$
The de-Broglie wavelength of the proton is given by
$${\lambda _i} = {h \over {{p_i}}} = {h \over {\sqrt {2{m_p}{K_i}} }} = {h \over e}\sqrt {{{4\pi {\varepsilon _0}a} \over {240{m_p}}}} $$
$$ = 4.2 	imes {10^{ - 15}}{\left( {{{10 	imes {{10}^{ - 15}}} \over {9 	imes {{10}^9} 	imes 240 	imes (5/3) 	imes {{10}^{ - 27}}}}} ight)^{1/2}}$$
= 7 fm.

Dual Nature of Matter and Radiation

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