If potential energy between a proton and an electron is given by |U|=k e2 / 2 R3, where e is the charge of electron and R is the radius of atom, then radius of Bohr's orbit is given by (h= Planck's constant, k= constant )

Question

If potential energy between a proton and an electron is given by |U|=k e2 / 2 R3, where e is the charge of electron and R is the radius of atom, then radius of Bohr's orbit is given by (h= Planck's constant, k= constant ) (A) (k e2 m/h2) (B) (6 π2/n2) (k e2 m/h2) (C) (2 π/n) (k e2 m/h2) (D) (4 π2 k e2 m/n2 h2)

Tardigrade Admin · Accepted Answer

U=-(k e2/2 R3), F=-(d U/d R)=-(3 k e2/2 R4) But, F=(m v2/R) ⇒ (m v2/R)=(3 k e2/2 R4) Also, m v R=(n h/2 π) Solve to get: R=(6 π2 k e2 m/n2 h2)

Q. If potential energy between a proton and an electron is given by $∣ U ∣ = k e^{2} /2 R^{3}$ , where $e$ is the charge of electron and $R$ is the radius of atom, then radius of Bohr's orbit is given by $(h =$ Planck's constant, $k =$ constant $)$

$\frac{k e ^{2} m}{h ^{2}}$

$\frac{6 π ^{2}}{n ^{2}} \frac{k e ^{2} m}{h ^{2}}$

$\frac{2 π}{n} \frac{k e ^{2} m}{h ^{2}}$

$\frac{4 π ^{2} k e ^{2} m}{n ^{2} h ^{2}}$

$U = - \frac{k e ^{2}}{2 R ^{3}},$
$F = - \frac{d U}{d R} = - \frac{3 k e ^{2}}{2 R ^{4}}$
But, $F = \frac{m v ^{2}}{R}$
$\Rightarrow \frac{m v ^{2}}{R} = \frac{3 k e ^{2}}{2 R ^{4}}$
Also, $m v R = \frac{nh}{2 π}$
Solve to get: $R = \frac{6 π ^{2} k e ^{2} m}{n ^{2} h ^{2}}$

Q. If potential energy between a proton and an electron is given by ∣U∣=ke2/2R3, where e is the charge of electron and R is the radius of atom, then radius of Bohr's orbit is given by (h= Planck's constant, k= constant )

h2ke2m​

n26π2​h2ke2m​

n2π​h2ke2m​

n2h24π2ke2m​