Question

Write the expression for calculating the radius of an orbit in hydrogen atom

Answer 1

Bohr Radius of a H atom of the first shell =5.29∗10−11m

= 5.29 *10^(-11) m * 100 cm / m ; Cancelling the m and multiplying ,

Bohr Radius =5.29∗10−9cm.

Hope that helps !

If you wish to know the derivation , proceed .

Derivation.

Let

Mass of electron = m=9.109384∗10−31kilogram;

Velocity of electron = v;

Charge of Electron = Charge of Proton =e=1.6021766∗10−19coulomb ;

Bohr Radius = R;

Planck’s Constant =h=6.62607∗10−34Joulesecond;

Coulumb’s Constant =K=8.9875518∗109Newtonm2/(C2) ;

Angular Momentum = L;

Shell number = n = 1 ( Since this is a Hydrogen Atom ) ;

Let us assume that the electron orbits in a perfectly circular orbit around the nucleus.

Then Centrifugal force =m∗v2R.

Coulumb’s Force = K×e×eR2.

Since Centrifugal force tries to push the electron away from the centre and Coulumb’s Force tries to pull it towards the Centre and we know that there is no net change in Radius ,

Centrifugal Force = Coulumb’s Force

mv2R=Ke2R2 Cancelling one of the R ’s ,

mv2=Ke2R.

Now , the angular momentum L is defined as cross product of mass of electron and its momentum (p).

Thus ,

L = R x p = R×p.

Since p vector is direction of electron , p vector and radius are perpendicular;

Thus ,

L=R×p=R∗(m∗v)=R∗m∗v;

Bohr assumed that the Angular momentum is quantized as

L=(shellnumber)∗(Planck′sConstant)/(2∗π)

L=R∗m∗v=nh2π

Rearranging ,

v=nh2πRm.

Let us plug in this value of v in our earlier equation

mv2=Ke2R.

Thus ,

m∗[nh2∗π∗R∗m]2=Ke2R.

m∗n2h22∗π∗m)21R2=Ke2R.

Cancelling one of the R ’s again

m∗n2h2(2∗π∗m)21R=Ke2.

Transposing the R ,

R = m∗n2∗h2(2∗π∗m)2Ke2

Cancelling one of the m ’s ,

R = n2∗h2(2π)2(K∗e2)m

Taking the values of the constants from a calculator and plugging in the values ,

We get R = 5.2909716×10−11m.

= 5.2909716×10(−11)×100cm

Thus ,

R = 5.2909716×10−9cm = 5.2909716×10−11m.