Question

An alpha particle after passing through a potential difference of $2\times {10}^6$ volt falls on a silver foil. The atomic number of silver is 47. The K.E. of the $\alpha$-particle at the time of falling on the foil is:

• A. $6.4\times {10}^{-13}J$
• B. $4.5\times {10}^{-13}J$
• C. $6.8\times {10}^{-13}J$
• D. $5.8\times {10}^{-13}J$

Answer 1

Answer: (A)

$6.4\times {10}^{-13}J$

Kinetic energy of $\alpha$-particle is given by :

$KE=eV$

$\alpha$-particle has 2 electrons.

$KE=2\times 2\times {10}^{6}eV=4\times {10}^6\times 1.602\times {10}^{-19}J=6.4\times {10}^{-13}J$