Question

A proton is fired from very far away towards a nucleus with charge Q = 120 e, 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, $m_p=\left(5/3\right)\times {10}^{-27}kg$ and

$h/e=4.2\times {10}^{-15}J.s/C;\frac{1}{4\pi {ϵ}_0}=9\times {10}^{9}m/F;1fm={10}^{-15}m$)

• A. $7fm$
• B. $8fm$
• C. $9fm$
• D. $10fm$

Answer 1

Answer: (A)

$7fm$

Given here,

$q_1=120\times e$

$q_2=e$ ( charge on proton)

It is based on conservation of energy of proton.

$\frac{1}{2}m{v}^2=k\frac{q_1{q}_2}{R}$

$v=\sqrt{\frac{2K{q}_1{q}_2}{mR}}$

Also, we know, de’broglie wavelength,

$\lambda =\frac{h}{mv}$

$\lambda =\frac{h\sqrt{R}}{\sqrt{2mK{q}_1{q}_2}}$

Thus,

$\lambda =\frac{h\sqrt{10\times {10}^{-15}}}{\sqrt{2\times \frac{5}{3}\times {10}^{-27}\times 9\times {10}^9\times 120\times e^2}}$

$\lambda =\frac{\frac{h}{e}\times {10}^{-7}}{6\times {10}^{-8}}$

$\lambda =\frac{4.2\times {10}^{-15}\times {10}^{-7}}{6\times {10}^{-8}}$

$\lambda =7\times {10}^{-15}m=7fm$

Option A is correct.