Question

An $\alpha -$particle passes through a potential difference of $2\times {10}^6$ volt and then it becomes incident on a silver foil. The charge number of silver is $47$, the kinetic energy of $\alpha -$ particles at a distance $5\times {10}^{-14}$m from the nucleus will be (in joules).

• A. $6.4\times {10}^{-13}$
• B. $4.3\times {10}^{-13}$
• C. $2.1\times {10}^{-13}$
• D. $3.4\times {10}^{-14}$

Answer 1

The correct option is C $2.1\times {10}^{-13}$

total energy=potential energy + kinetic energy

W = - (Z q² / 4 p e_o r ) + (Z q² / 8 p e_or ) = - (Z q² / 8 p e_o r)

Where

Z is the atomic number (=1 for Hydrogen)

q is the charge of an electron=(1.6 x 10^{- 9}coulomb)

And e _o is the permittivity of free space=(8.854 x 10^{- 12}farad / meter)

after solving there equation we get,

total energy= -13.6 eV

kinetic energy= +13.6 eV

potential energy= -26.6 eV