## A particle of mass m moves in circular orbits with potential energy V(r) = F r , where F is positive constant ….

Q: A particle of mass m moves in circular orbits with potential energy V(r) = F r , where F is positive constant and r is its distance from origin . Its energy are calculated using the Bohr model . If radius of the particle’s orbit is denoted by R and its speed and energy are denoted by v and E respectively , then for the n^th orbit (here h is the Planck’s constant)

(a) $R \propto n^{1/3}$ and $v \propto n^{2/3}$

(b) $R \propto n^{2/3}$ and $v \propto n^{1/3}$

(c) $E = \frac{3}{2} (\frac{n^2 h^2 F^2}{4\pi^2 m})^{1/3}$

(d) $E = 2 (\frac{n^2 h^2 F^2}{4\pi^2 m})^{1/3}$

Click to See Solution :
Ans: (b , c)
Sol: Force $|F| = |-\frac{dV}{dr}|$

$F = \frac{m v^2}{r}$ …(i)

$m v r = \frac{n h}{2 \pi}$ ….(ii)

From (i) & (ii)

$F = \frac{m}{r} (\frac{n h}{2 \pi m r})^2$

$F = \frac{m n^2 h^2}{4 \pi^2 m^2 r^3}$

$F = \frac{ n^2 h^2}{4 \pi^2 m r^3}$

Since F = constant

$r^3 \propto n^2$

$r \propto n^{2/3}$

$v = \frac{n h}{2 \pi m r}$

$v \propto \frac{n}{n^{2/3}}$

$v \propto n^{1/3}$

$K.E = \frac{1}{2}m v^2 = \frac{F r}{2}$

Total Energy = P.E + K.E

$T.E = F r + \frac{F r}{2}$

$T.E = \frac{3 F r}{2}$

$T.E = E = \frac{3F}{2} \times (\frac{n^2 h^2}{4 \pi^2 r n F})^{1/3}$

$E = \frac{3}{2} (\frac{n^2 h^2 F^2}{4 \pi^2 m})^{1/3}$

## The Bohr model for the spectra of a H-atom

Q: The Bohr model for the spectra of a H-atom

(a) will not be applicable to hydrogen in the molecular from

(b) will not be applicable as it is for a He-atom

(c) is valid only at room temperature

(d) predicts continuous as well as discrete spectral lines

Ans: (a) & (b)

## A set of atoms in an excited state decays

Q: A set of atoms in an excited state decays

(a) in general to any of the states with lower energy

(b) into a lower state only when excited by an external electric field

(c) all together simultaneously into a lower state

(d) to emit photons only when they collide

Ans: (a)

## For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence ….

Q: For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

(a) because Bohr model gives incorrect values of angular momentum

(b) because only one of these would have a minimum energy

(c) angular momentum must be in the direction of spin of electron

(d) because electrons go around only in horizontal orbits

Ans: (a)