JEE Advance - Physics (2012 - Paper 1 Offline - No. 18)

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 $$\to$$ $$\infty$$), its potential energy is

$${U_i} = \mathop {\lim }\limits_{r \to \infty } {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over r} = 0$$.

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

$${\lambda _i} = {h \over {{p_i}}} = {h \over {\sqrt {2{m_p}{K_i}} }} = {h \over e}\sqrt {{{4\pi {\varepsilon _0}a} \over {240{m_p}}}} $$

$$ = 4.2 \times {10^{ - 15}}{\left( {{{10 \times {{10}^{ - 15}}} \over {9 \times {{10}^9} \times 240 \times (5/3) \times {{10}^{ - 27}}}}} \right)^{1/2}}$$

= 7 fm.