A $1 M e V$ proton is sent against a gold leaf $(Z = 79)$ . Calculate the distance of closest approach for head-on collision.

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Solution

$1.137 \times 10^{- 13} m$ .
$d = \frac{{Z e}^{2}}{4 π ε_{0} (\frac{1}{2} m v^{2})}$ .

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A 1MeV proton is sent against a gold leaf (Z=79). Calculate the distance of closest approach for head-on collision.

1.137×10−13m.d=Ze24πε0(12mv2).

A $1 M e V$ proton is sent against a gold leaf $(Z = 79)$ . Calculate the distance of closest approach for head-on collision.

$1.137 \times 10^{- 13} m$ .
$d = \frac{{Z e}^{2}}{4 π ε_{0} (\frac{1}{2} m v^{2})}$ .