Question

the nucleus of hydrogen atom is a sphere of radius 10 to the power minus 15 metre. the electric field at the surface of the nucleus is?

Answer 1

electric field at the surface of the nucleus is 1.44 × 10²¹ N/C

hydrogen atom has one electron revolves around the nucleus and one proton on the nucleus.

so, electric field intensity due to proton at the surface of nucleus, E = kq/r²

where , q = 1.6 × 10^-19 C

and r = 10^-15 m

so, E = 9 × 10^9 × 1.6 × 10^-19/(10^-15)²

= 9 × 1.6 × 10^-10/(10^-30)

= 14.4 × 10^20 N/C

= 1.44 × 10²¹ N/C

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Answer 2

$\huge\bold\purple{Answer:-}$

hydrogen atom has one electron revolves around the nucleus and one proton on the nucleus.

so, electric field intensity due to proton at the surface of nucleus, E = kq/r²

where , q = 1.6 × 10^-19 C

and r = 10^-15 m

so, E = 9 × 10^9 × 1.6 × 10^-19/(10^-15)²

= 9 × 1.6 × 10^-10/(10^-30)

= 14.4 × 10^20 N/C

= 1.44 × 10²¹ N/C