The closest distance of approach of an Alpha particle travelling with a velocity V towards a stationary nucleus is d for the closest distanc
Answer (1)
16th Mar, 2020
Hello Student , this can be easily derived as shown below
Let m and v be the mass and velocity of alpha particle directed towards the centre of nucleus.
Kinetic Energy of alpha particle = K = 1/2 * mv²
Positive charge on the nucleus is Ze and charge of alpha particle is 2e
Electrostatic potential energy of the alpha particle when it is at a distance d from the centre of nucleus is U = 2Ze²/4πDε 0
At closest approach Kinetic Energy of particle = Potential Energy
So >>> 1/2 * mv² = 2Ze²/4πDε 0
D = 4Ze²/mv²4πε 0
or D = Ze²/mv²πε 0
Hope my answer was helpful. Do not memorize the formula remember the derivation using concept.
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