A proton is fired from very far away towards a nucleus with charge , where 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, ; ;

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, $m_{p} = (\frac{5}{3}) \times 10^{- 27} kg$ ; $\frac{h}{e} = 4.2 \times 10^{- 15} Js / C$ ; $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} m / F$ ; $1 fm = 10^{- 15} m$ ) [2012]

Smart Achievers · Accepted Answer

(7)

From energy conservation, Loss in K.E. of proton = Gain in potential energy of the proton-nucleus system

$\frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$ $∴$ $\frac{p^{2}}{2 m} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$

$∴$ $\frac{1}{2 m} (\frac{h^{2}}{λ^{2}}) = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$ $∴$ $λ = \sqrt{\frac{4 π ε_{0} r . h^{2}}{q_{1} q_{2} (2 m)}}$

Putting the values of $4 π ε_{0}$ , $r$ , $h$ , $q_{1}$ , $q_{2}$ , and $m$ , we get the de-Broglie wavelength of proton, $λ = 7 fm$

Solution

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