An alpha particle of mass 6.64x10^-27 kg moving at 1.6x10^7 m s^-1 approaches a stationary gold nucleus (Z=79) head-on. Treat the nucleus as fixed. Use 1/(4piepsilon_0)=9.0x10^9 N m^2 C^-2 and e=1.6x10^-19 C. Find the distance of closest approach in fm (1 fm=10^-15 m), to one decimal.

JEE Advanced levelNumerical answerDifficulty 3 of 3as043

An alpha particle of mass $6.64 \times 1 0^{- 27} kg$ moving at $1.6 \times 1 0^{7} m s^{- 1}$ approaches a stationary gold nucleus ( $Z = 79$ ) head-on. Treat the nucleus as fixed. Use $1/ (4 π ϵ_{0}) = 9.0 \times 1 0^{9} N m^{2} C^{- 2}$ and $e = 1.6 \times 1 0^{- 19} C$ . Find the distance of closest approach in fm ( $1 fm = 1 0^{- 15} m$ ), to one decimal.

Numerical answer type, answer in fm, tolerance 0.2.

K =

(1/2) m v^{2}

= 0.5 x 6.64e-27 x

(1.6 e 7)^{2}

= 8.4992e-13 J. At closest approach K = k(2e)(79e)/d, so d = 9e9 x 2 x 79 x

(1.6 e - 19)^{2} /8.4992 e - 13

= 4.2831e-14 m = 42.8 fm.