Question

Consider an electron in a hydrogen atom in the n=2,l=0 state. At what radius ( in units of a0) is the electron most likely to be found?

Answer 1

Given,

n = 2

l = 0

i.e electron is in 2s state.

Atomic number, Z = 1

Now,

As we know,

Let the velocity of electron be v

let the radius of rotation be r

let the mass of electron be m

Total charge on nucleus is Z*e where Z is atomic number

for an electron to rotate around nucleus in a circular path, Electrostatic force between electron and nucleus should be equal to centrifugal force acting on the electron

=> k ((Z*e)*e) / r² = mv2/r            where k = (1 / (4₀))

=> k*Ze² = mv²r = (m*v*r)*v

As we know,

angular momentum of electron in nth orbit, m*v*r = n*h/2

=> (n*h/2)*v = k*Ze²

=> v = (k*Ze²) / (n*h/2) .......(1)

Now, We know that

=> m*v*r = n*h/2

=> r = (n*h/2) / (m*v)

Now put the value of v from equation(1)

=> r = (n*h/2) / (m((k*Ze²) / (n*h/2)))

=> r = (n*h/2)² / (m*k*Z*e²)

=> r =[(h/2)² / (m*k*e²)]* (n²/Z)

or r = a₀ * (n²/Z) Where a₀ = [(h/2)² / (m*k*e²)]

Thus,

radius , r = (n²/Z)*a₀

So,

r₁ = a₀*(1²/1)

= a₀

Similarly,

r₂ = a₀(2²/1)   

= (4/1) * a₀ = 4*a0

Now,