What is the distance of closest approach when a 5 MeV proton approaches a gold nucleus (Z=79) ?

Question

What is the distance of closest approach when a 5 MeV proton approaches a gold nucleus (Z=79) ? (A) 4 × 10-14 m (B) 2 × 10-14 m (C) 3 × 10-14 m (D) None of these

Tardigrade Admin · Accepted Answer

E =(( Ze )( e )/4 π ε o r )=(9 × 109 × 79 ×(1.6 × 10-19)2/ r × 1.6 × 10-13) r=2.3 × 10-14 m

Q. What is the distance of closest approach when a $5 M e V$ proton approaches a gold nucleus $(Z = 79)?$

$4 \times 1 0^{- 14} m$

$2 \times 1 0^{- 14} m$

$3 \times 1 0^{- 14} m$

None of these

$E = \frac{( Z e ) ( e )}{4 π ε _{o} r} = \frac{9 \times 1 0 ^{9} \times 79 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{r \times 1.6 \times 1 0 ^{- 13}}$
$r = 2.3 \times 1 0^{- 14} m$

Q. What is the distance of closest approach when a 5MeV proton approaches a gold nucleus (Z=79)?

4×10−14m

2×10−14m

3×10−14m