An alpha-particle of energy 5 MeV is scattered through 180-0- by a fixed uranium nucleus- The closest distance is in the order of1A-0-10-12- cm10-10- cm10-16- cm

Question

Toppr Admin · Accepted Answer

The distance of closest approach isd-Ze24-x3C0-x3F5-0EkHere- Z - 92- Ek-5MeV-5-xD7-106-xD7-1-6-xD7-10-x2212-19JHence-d-9-xD7-109-xD7-2-xD7-92-1-6-xD7-10-x2212-19-25-xD7-106-xD7-1-6-xD7-10-x2212-19

An $α$ -particle of energy $5 M e V$ is scattered through $180^{0}$ by a fixed uranium nucleus. The closest distance is in the order of
$1$ $A^{0}$
$10^{- 12}$ cm
$10^{- 10}$ cm
$10^{- 16}$ cm

The distance of closest approach is
$d = \frac{Z e^{2}}{4 π ϵ_{0} E_{k}}$
Here, Z = 92,
$E_{k} = 5 M e V = 5 \times 10^{6} \times 1.6 \times 10^{- 19} J$
Hence,
$d = \frac{9 \times 10^{9} \times 2 \times 92 (1.6 \times 10^{- 19})^{2}}{5 \times 10^{6} \times 1.6 \times 10^{- 19}}$

An α-particle of energy 5MeV is scattered through 1800 by a fixed uranium nucleus. The closest distance is in the order of1A010−12 cm10−10 cm10−16 cm

The distance of closest approach isd=Ze24πϵ0EkHere, Z = 92, Ek=5MeV=5×106×1.6×10−19JHence,d=9×109×2×92(1.6×10−19)25×106×1.6×10−19

An $α$ -particle of energy $5 M e V$ is scattered through $180^{0}$ by a fixed uranium nucleus. The closest distance is in the order of
$1$ $A^{0}$
$10^{- 12}$ cm
$10^{- 10}$ cm
$10^{- 16}$ cm

The distance of closest approach is
$d = \frac{Z e^{2}}{4 π ϵ_{0} E_{k}}$
Here, Z = 92,
$E_{k} = 5 M e V = 5 \times 10^{6} \times 1.6 \times 10^{- 19} J$
Hence,
$d = \frac{9 \times 10^{9} \times 2 \times 92 (1.6 \times 10^{- 19})^{2}}{5 \times 10^{6} \times 1.6 \times 10^{- 19}}$