SATHEE: Modern Physics 4 Question 25

Modern Physics 4 Question 25

26. 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 $10 f m$ to the nucleus. The de-Broglie wavelength (in units of $f m$ ) of the proton at its start is [Take the proton mass, $m_{p} = (5 / 3) \times 10^{- 27} k g; h / e = 4.2 \times 10^{- 15}$ $J - s / C$ ;

$\frac{1}{4 π ε_{0}} = 9 \times 10^{9} m / F; 1 f m = 10^{- 15} m]$

(2013 Adv.)

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Solution:

$r =$ closest distance $= 10 f m$ .

From energy conservation, we have

$\begin{aligned} K_{i} + U_{i} & = K_{f} + U_{f} \\ or \\ or & = 0 + \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r} \\ K & = \frac{1}{4 π ε_{0}} \cdot \frac{(120 e) (e)}{r} \end{aligned}$

de-Broglie wavelength

$λ = \frac{h}{\sqrt{2 K m}}$

Substituting the given values in above two equations, we get

$λ = 7 \times 10^{- 15} m = 7 f m$

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Modern Physics 4 Question 25

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