Answer :

Given:

Radius of nucleus=10^{-15}m

Radius of 1-s charge cloud=1.3× 10^{-11}m=r_{1}

Radius of 2-s charge cloud =5.2× 10^{-11}m=r_{2}

(a)for calculating electric field at a point just inside 1s cloud

Consider a gaussian spherical surface of radius just equal to the radius of 1s cloud=r_{1}

By symmetry all points on this surface are equivalent and electric field at all points have same magnitude and is in radial direction

Therefore flux passing through this surface is given by

…(i)

We know that,

By Gauss’s law, flux of net electric field (E⃗ ) through a closed surface S equals the net charge enclosed (q_{in}) by the surface divided by ϵ_{0}

Charge enclosed by this sphere =total charge on the nucleus=q

Now,

q= C

**C**

Using gauss’s law and eqn.(i)

Putting the values of r_{1} ,q and ϵ_{0}

N/C

**N/C**

**Therefore electric field at a point just inside the 1s cloud is given by 3.4× 10 ^{13}N/C**

(b)for calculating electric field at a point just inside the 2s cloud

Consider a gaussian spherical surface of radius just equal to the radius of 2s cloud=r_{2}

By symmetry all points on this surface are equivalent and electric field at all points have same magnitude and is in radial direction

Therefore flux passing through this surface is given by

…(i)

We know that,

By Gauss’s law, flux of net electric field (E⃗ ) through a closed surface S equals the net charge enclosed (q_{in}) by the surface divided by ϵ_{0}

Charge enclosed by this sphere =total charge on the nucleus+ charge of 2 1s electrons=q

Now,

q= C

**C**

Using gauss’s law and eqn.(i)

Putting the values of r_{1} ,q and ϵ_{0}

N/C

**N/C**

**Therefore electric field at a point just inside the 2s cloud is given by 1.1× 10 ^{12}N/C**

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