Quiz 3: Modern Physics Solution

Max. marks:10 Time Allowed:20 minutes

Name: Roll no:

1. An atom has two energy levels with a transition wavelength of 580 nm. At room temperature 4 × 1020 atoms are in the lower state. (a) How many atoms occupy the upper state, under conditions of thermal equilibrium? (b) Suppose instead that 7×1020 atoms are pumped into the upper state, with 4×1020 in the lower state. This is clearly a non-equilibrium state. How much energy in Joules could be released in a single pulse of light as equilibrium is restored? −23 The Boltzmann factor is exp(−E/kBT ) where kB = 1.38 × 10 J/K. Answer 1:

If Nb is the higher energy state and Na is the lower energy state, then the energy diﬀerence between these two states can be calculated using transition wavelength of 5800 A.˚ i.e., hc E − E = hf = b a λ (6.63 × 10−34 Jsec)(3 × 108 m/sec) = 5800 × 10−10 m −19 Eb − Ea = 3.43 × 10 J

−28 kBT = 1.38 × 10 J/K × 300 K = 4.1 × 10−21 J.

According to the Boltzmann distribution, the population in any state i is,

−Ei/kB T Ni = e .

Therefore, the ratio of Na and Nb in equilibrium is, eq N −19 −21 b −(Eb−Ea)/kB T −(3.43×10 J/4.1×10 J) eq = e = e Na = 1.1 × 10−36 eq eq × × −36 Nb = Na 1.1 10 = (4 × 1020)(1.1 × 10−36)

= 4.4 × 10−16,

Date: 8 March, 2013 1 Max. marks:10 Time Allowed:20 minutes

which is very small. This shows that the energy gap between the given states Na and

Nb, at room temperature, is so large that almost all the electrons are present in lower

state Na. There is hardly any electrons in the upper state.

(b) When atoms are pumped in the upper state, this is a non-equilibrium situation. Energy is released up till equilibrium is restored. Therefore,

non-eq × 20 Na = 4 10 non-eq × 20 Nb = 7 10 non-eq non-eq non-eq N = Na + Nb = 11 × 1020.

We are interested in number of atoms which restores the equilibrium. Since the ratio, −36 Na/Nb = 1.1 × 10 , remains the same (the gap width does not change), therefore,

non-eq non-eq × 20 Na + Nb = 11 10 non-eq × 35 × 20 Nb (1 + 9.09 10 ) = 7 10 non-eq × −15 Nb = 1.21 10 .

The number of atoms which contribute to restore equilibrium is,

non-eq − non-eq ∆N = N Nb ≈ 7 × 1020.

The energy released in a single pulse is thus,

hc E = ∆N × λ = 7 × 1020 × 3.43 × 10−19 J

= 240 J.

2. Suppose a proton and an electron were held together in a hydrogen atom by gravitational forces only. (a) Find the formula for the energy levels of such an atom, and the radius of the ground state Bohr orbit.

Date: 8 March, 2013 2 Max. marks:10 Time Allowed:20 minutes

(b) Find the numerical value of the Bohr radius for this “gravitational” atom. Reduced Planck’s constant ~ = 1.06 × 10−34 Js, Mass of proton M = 1.67 × 10−27 kg, Mass of electron m = 9.11 × 10−31 kg, Gravitational constant G = 6.67 × 10−11 Nm2/kg2. Answer 2: Ler r be the radius of electron’s orbit, then gravitational force between proton and electron is, mM F = G . (1) g r2 Centripetal force needed to keep the electron in a circular orbit is provided by this gravitational force. If v is the orbital speed of electron, centripetal force will be, mv2 F = . (2) c r Comparing equations (1) and (2), we obtain, mv2 mM = G r r2 M v2 = G . (3) r According to Bohr atomic model “the allowed orbits are those for which the electron’s orbital angular momentum about the nucleus is an integral multiple of ~”, i.e.,

mvr = n~ n = 1, 2, 3, ..... n~ ⇒ v = mr n2~2 v2 = . (4) m2r2 Comparing equations (3) and (4), GM n2~2 = r m2r2 ~2 ⇒ r = n2 , GMm2 which is the required formula for the radius of the Bohr orbit. For the ground state, set n = 1. Total energy of the atom, which consists of both the kinetic and potential energy terms is, 1 Mm E = mv2 − G . 2 r

Date: 8 March, 2013 3 Max. marks:10 Time Allowed:20 minutes

Substitute value of v2 from equation (3), ( ) 1 M Mm E = m G − G 2 r r GMm 1 = − · 2 r GMm 1 = − · 2 2 ~2 n GMm2 GMm GMm2 = − · 2 n2~2 M 2m3 E = −G2 , 2n2~2

which is the required formula for energy levels. (b) Since,

~2 r = n2 n GMm2

For n = 1,

~2 r = . 1 GMm2

Substitution of the given values yields,

(1.06 × 10−34 Js)2 r = 1 6.67 × 10−11 Nm2/kg2 × 1.67 × 10−27 kg × (9.11 × 10−31 kg)2 = 1.2 × 1029 m.

Note r1 is a huge value, because gravitational attraction is much much weaker than electrostatic interaction. Even though this is not required, we calculate the energy in the lowest orbit. Given

M 2m3 E = −G2 , n 2n2~2

and setting n = 1,

M 2m3 E = −G2 1 2~2 (1.67 × 10−27 kg)2 × (9.11 × 10−31 kg)3 = −(6.67 × 10−11 Nm2/kg2)2 × 2(1.06 × 10−34 Js)2 = −4.17 × 10−97 J.

Date: 8 March, 2013 4