Discussion Forum

1)

A proton is fired from very far away towards a nucleus with charge Q=120 e, where e is the electronic charge. It makes the 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,

$m_{p}=(5/3)\times 10^{-27}$kg

$h/e=4.2 \times 10^{-15}J-s/C$

$\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} m/F$; 1 fm= $10^{-15}$m]

Answer:

Option C

Explanation:

 r= closet distance =10fm

 from energy conservation , we have

 $K_{i}+U_{i}=K_{f}+U_{f}$

 or   $K+0=0+ \frac{1}{4\pi \epsilon_{0}}.\frac{q_{1}q_{2}}{r}$ 

   $K= \frac{1}{4\pi \epsilon_{0}}.\frac{(120e)(e)}{r}$.........(i)

 de-Broglie wavelength

 $\lambda=\frac{h}{\sqrt{2Km}}$...........(ii)

Substituting the given values in above two equations, we get

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