AP EAPCET 19-Aug-2021 Shift 1 Paper

Given, potential energy between proton and electron,
U=

Ke2

3R3

As we know that,
Force, F=

−dU

dR

∴F=−

Ke2

3

d

dR

R−3
⇒

mv2

R

=−

Ke2

3

,(−3)R−4
where, m is mass, v is speed and R is radius.
∴

mv2

R

=

Ke2

R4

⇒R3=

Ke2

mv2

...(i)
From uncertainty principle,
L=mvR=

nh

2π

⇒v=

nh

2π‌mR

Substituting the value in Eq. (i), we get
R3=

Ke2

m. n2h2 4π2m2R2

R3=

Ke2.4π2m2R2

mn2h2

∴R=

4π2.KR2.m

n2h2