Question

Find electric field intensity at a distance 2A from gold nucleus having Z=79

Answer 1

electric field intensity would be 1.044 × 10¹² N/C.

we have to find the electric field intensity at a distance 2Å from gold nucleus having Z = 29.

using formula, E = Kq/r²

where r = 2Å = 2 × 10^-10 m, Charge, q = 29 e¯

= 29 × 1.6 × 10^-19 C = 4.64 × 10^-18C

now E = (9 × 10^9 × 4.64 × 10^-18)/(2 × 10^-10)²

= (41.76 × 10^-9)/(4 × 10^-20)

= 10.44 × 10¹¹ N/C

= 1.044 × 10¹² N/C

Therefore, electric field intensity would be 1.044 × 10¹² N/C.

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