Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is(a) 14.4 Z/V Å(b) 14.4 V/Z Å(c) 1.44 Z/V Å(d) 1.44 V/Z Å
Question
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is(a) 14.4 Z/V Å(b) 14.4 V/Z Å(c) 1.44 Z/V Å(d) 1.44 V/Z Å
Solution
The distance of closest approach (r) of an alpha particle to the nucleus can be calculated using the formula:
r = (2 * K * e^2 * Z) / (V * e)
where:
- K is Coulomb's constant (9 x 10^9 N.m^2/C^2)
- e is the charge of the alpha particle (1.6 x 10^-19 C)
- Z is the atomic number of the nucleus
- V is the potential difference in volts
Substituting the values of K and e in the formula, we get:
r = (2 * 9 x 10^9 * (1.6 x 10^-19)^2 * Z) / (V * 1.6 x 10^-19)
Simplifying the above expression, we get:
r = (2.88 x 10^-14 * Z) / V
Converting the distance from meters to Angstroms (1 Angstrom = 10^-10 meters), we get:
r = (2.88 x 10^-14 * Z / V) * 10^10
r = 14.4 Z/V Å
So, the correct answer is (a) 14.4 Z/V Å.
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