Question

An $\alpha -$particle passes through a potential difference of $2\times {10}^{6}V$ and then it becomes incident on a silver foil. The charge number of silver is $47$. The energy of incident particles will be (in joules)

• A. $5\times {10}^{-12}$
• B. $6.4\times {10}^{-13}$
• C. $5.8\times {10}^{-14}$
• D. $9.1\times {10}^{-15}$

Answer 1

The correct option is B $6.4\times {10}^{-13}$

$E=qV$
$=2\times 1.6\times {10}^{-19}\times 2\times {10}^6$
$=6.4\times {10}^{-13}$ joules