Two free response questions were posted on ^{th} March 2008

**Free Response Question No.1**

An electron is accelerated by a positive potential *V* of the order of a few volts.

(a) Derive an expression for the de Broglie wave length *λ* of the electron in terms of the electron mass *m*, electron charge *e*, Planck’s constant *h* and the accelerating potential *V*.

(b) A photon and an electron have the same wave length *λ. *Determine their kinetic energy in term of the wave length.

(c) Explain with reason, which particle in (b) has greater kinetic energy.

(d) If you substitute the known values of the electron mass *m*, electron charge *e* and the Planck’s constant *h*, the de Broglie wave length of an electron for accelerating voltages less than 50 volts or so will work out to be very nearly equal to √(150/*V*) Ǻ where *V* is the accelerating voltage (in volt). Using this simple expression, calculate the de Broglie wave length of an electron of energy 15 electron volt.

(e) Explain why you cannot use the simple expression in (d) for calculating the de Broglie wave length if the accelerating voltage is high as for example, in an X-ray tube.

(a) .The electron gains kinetic energy*Vq*=

*Ve*so that we can write,

*Ve* = *p*^{2}/2*m* where *p* is the momentum of the electron. From this equation,

*p* = √(2*Vem*).

The de Broglie wave length *λ* of the electron is given by

*λ* = *h/p* = *h/*√(2*Vem*) where *h *is Planck’s constant.

(b) The energy of a photon is fully kinetic and is *hν* where the frequency *ν *=* c*/*λ *where *c *is the speed of light.* *Therefore, the kinetic energy of the photon is *hc/λ*.

The kinetic energy of the electron is *p*^{2}/2*m* where *p* is its momentum given by

*p =h*/*λ*

Therefore, kinetic energy of the electron = *h*^{2}/2*λ*^{2}*m*

(c) Since the photon and the electron have the same wave length, they have the *same* *momentum* (in accordance with de Broglie’s relation *p = h*/*λ*). The photon has speed *c* which is the maximum speed attainable. An electron can never attain this speed since its rest mass is not zero (unlike the photon). The mass of the electron is therefore greater than the dynamic mass of the photon of the same wave length.

At non-relativistic speeds, the kinetic energy of a particle is *p*^{2}/2*m. *Therefore, when the momentum is the same, the heavier particle electron has smaller kinetic energy.

[This part of the question will prompt you to think of the situation when the wave length is extremely small so that the speed of the electron approaches the speed of light. (In the case of an electron of energy 2 MeV, the speed will be 2.93×10^{8} ms^{–1}** **and the** **mass will be 4.36 ×10^{ –30} kg where as the rest mass of the electron is 0.91×10^{ –30} kg only). When you consider the relativistic mass increase, the kinetic energy of any particle is (*m *– *m*_{0})c^{2} where *m* is the mass of the particle while moving and *m*_{0} is the rest mass. Since the rest mass of the photon is zero, its kinetic energy is mc^{2}. You can continue by calculating the mass of the electron while moving, using the well known expression *m = m*_{0}/√(1 – v^{2}/*c*^{2}) and hence the kinetic energy of the electron. You will find that the kinetic energy of the photon is greater than that of the electron of the same wave length].

(d) An electron will have kinetic energy 15 eV if it is accelerated by a potential of 15 volts. Therefore, its wave length is given by

*λ* = √(150/*V*) Ǻ = √(150/15) Ǻ = 3.162 Ǻ

(e) At small accelerating voltages the velocity acquired by the electron is not comparable to the velocity of light. But when the accelerating voltage is large, the electron moves with speeds comparable to the speed of light and the relativistic mass increase is significant. The relation, *λ* = √(150/*V*) Ǻ is obtained by taking the mass of the electron to be constant. So, it cannot hold at large accelerating voltages.

**Free Response Question No.2**

**(a)** Derive an expression for the kinetic energy of the electron in a hydrogen atom in terms of the radius ‘*r*’ of the orbit, electronic charge ‘*e*’ and the permittivity ‘*ε*_{0}’_{ }of free space. Show that the potential energy of the electron is negative and of value twice that of the kinetic energy.

**(b)** The total energy *E*_{n} of the electron in the *n*^{th} orbit in a hydrogen atom is given by *E*_{n }= – *me*^{4}/8*ε*_{0}^{2}*n*^{2}*h*^{2} where ‘*m*’ is the mass of the electron (very nearly) and ‘*h*’ is Planck’s constant. Explain how this expression will get modified in *positronium* in which a positron is to be considered in place of the proton in a hydrogen atom.

**(c)** An electron and a positron moving along a straight line in opposite directions with equal speeds undergo a head-on collision and get annihilated, producing two photons of the same energy. Explain why there must be two photons produced instead of just one.

**(d)** Explain why the two photons produced in the annihilation stated in (c) must be of the same energy.

**(a)** For a dynamically stable orbit in a hydrogen atom the electrostatic attractive force between the electron and the nucleus (proton) supplies the centripetal force required for the circular motion of the electron around the nucleus so that we have

(1/4πε_{0})*e*^{2}/*r*^{2} = *m*v^{2}/*r*^{}

where ‘v’ is the velocity of the electron. From the above expression, the kinetic energy *K *of the electron is given by

*K* = ½ *m*v^{2} = *e*^{2}/8πε_{0}*r*

The potential energy *U *of the electron in the coulomb field of the nucleus is the product of the positive electric potential (produced by the nucleus) and the charge (–e) of the electron. Therefore,

*U *= [(1/4πε_{0})*e*/*r*](–e) = – *e*^{2}/4πε_{0}*r*

The potential energy of the electron is negative and of value twice that of the kinetic energy.

**(b)** In the case of a hydrogen atom the nucleus is a proton which is about 1840 times massive than the electron. In the expression for the total energy *E*_{n} of the electron in the *n*^{th} orbit in a hydrogen atom, given by *E*_{n }= – *me*^{4}/8*ε*_{0}^{2}*n*^{2}*h*^{2}, strictly speaking, the mass of the electron (*m*) is to be replaced by the *reduced mass* *μ* of the electron and the proton. Since the reduced mass *μ = mM/(m+M)* where *M *is the mass of the proton, it is almost equal to the electronic mass ‘*m*’.

But in positronium since the electron and the positron have the same mass *m*, the reduced mass is given by *μ *= *mm*/(*m*+*m*) = *m*/2. The expression for the total energy of the electron in positronium is therefore given by

*E*_{n }= – *me*^{4}/16*ε*_{0}^{2}*n*^{2}*h*^{2}

[Note that this is only half the value for the electron in a hydrogen atom].

**(c) **When an electron and a positron moving with equal speed in opposite directions undergo a head-on collision, the net momentum of the system is zero. The net momentum of the products produced in the process also must be zero in accordance with the law of conservation of momentum. A single photon will have its own momentum. Two photons carrying equal and opposite momentum ( in other words, two photons of the same wave length moving in opposite directions) can secure the condition of zero net momentum.

**(d)**Since the magnitude of the momentum of the photons is the same, their

*wave length*also

*must be the same*in accordance with de Broglie relation

*λ = h/p.*Since the energy of a photon is

*hc/λ*where

*h*is Planck’s constant and

*c*is the speed of light, the two photons must be of the same energy.

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