Friday, October 14, 2022

H C Verma solutions, ELECTRIC CURRENT THROUGH GASES, Chapter-41, Exercises, Q1 To Q10, Concepts of Physics, Part-II

Electric Current Through Gases


Exercises, Q1 to Q10


     1.  A discharge tube contains helium at low pressure. A large potential difference is applied across the tube. Consider a helium atom that has just been ionized due to the detachment of an atomic electron. Find the ratio of the distance traveled by the free electron to that by the positive ion in a short time dt after the ionization. 


ANSWER: Let the electric field near the ionized Helium atom be E. Due to the detachment of an atomic electron, there will be a positive charge equal to 'e' on the ion and a negative charge on the detached electron equal to '-e'. So both will experience an equal and opposite electric force,

F =eE

Let the mass of the electron =mₑ and mass of the ion =mₘ, then

Acceleration of the electron,

aₑ =Force/mass =eE/mₑ

Distance traveled by this electron in time dt,

S =½aₑ(dt)²

And the acceleration of the ion,

aₘ =eE/mₘ

So the distance traveled by the ion,

S' =½aₘ(dt)²

So the required ratio is

S/S' =aₑ/aₘ

  =(eE/mₑ)/(eE/mₘ)

  =mₘ/mₑ

  Now, mₘ =mass of two protons and two neutrons, neglecting the mass of the remaining electron.

        ≈4*1.67x10⁻²⁷ kg

        =6.68x10⁻²⁷ kg

and mₑ =9.1x10⁻³¹ kg

Hence the required ratio

S/S' =mₘ/mₑ

    =6.68x10⁻²⁷/9.1x10⁻³¹

    =66800/9.1

    =7340.   





 

     2.  A molecule of gas, filled in a discharge tube, gets ionized when an electron is detached from it. An electric field of 5.0 kV/m exists in the vicinity of the event. (a) Find the distance traveled by the free electron in 1 µs assuming no collision. (b) If the mean free path of the electron is 1.0 mm, estimate the time of transit of the free electron between successive collisions.    


ANSWER: (a) Force on the electron,

F =eE

Acceleration of the electron if its mass =m,

a =F/m =eE/m

The distance traveled by the electron in time t,

S = ut+½at²

  =0 +½(eE/m)t²

=½*(1.6x10⁻¹⁹*5000/9.1x10⁻³¹)*(10⁻⁶)²  m

=440 m.


(b) Let the time of transit of free electrons between successive collisions =t. From,

S =½at² 

Here, S =1.0 mm =0.001 m 

a =eE/m 

  =1.6x10⁻¹⁹*5000/9.1x10⁻³¹ m/s² 

 =8.79x10¹⁴ m/s² 

Hence,

0.001 =½*8.79x10¹⁴*t² 

→t² =0.002/(8.79x10¹⁴) 

     =2.28x10⁻¹⁸  

→t =1.50x10⁻⁹ s 

     =1.50 ns.         






 

     3.  The mean free path of electrons in the gas in a discharge tube is inversely proportional to the pressure inside it. The Crookes dark space occupies half the length of the discharge tube when the pressure is 0.02 mm of the mercury. Estimate the pressure at which the dark space will fill the whole tube. 


ANSWER: In the Crookes dark space the electrons emitted from the cathode travel a distance equal to the mean free path before they collide and cause the cathode to glow. So according to the question, the length of the Crookes dark space is inversely proportional to the pressure inside the given discharge tube. Let the length of the Crookes dark space = L and the pressure inside the discharge tube =p, then

L ∝ 1/p

So for the two cases when Crookes's dark spaces are L₁ and L₂ and the corresponding pressures are p₁ and p₂,

L₁ ∝ 1/p₁ and

L₂ ∝ 1/p₂. Hence

L₁/L₂ =p₂/p₁. 

Let the length of the discharge tube =l. Given, L₁ = l/2 and p₁ =0.02 mm of mercury, L₂ =l then p₂ =? 

So,

(l/2)/l =p₂/0.02

→p₂ =0.02/2 =0.01 mm of mercury.

         




 

     4.  Two discharge tubes have identical material structures and the same gas is filled in them. The length of one tube is 10 cm and that of the other tube is 20 cm. Sparking starts in both tubes when the potential difference between the anode and cathode is 100 V. If the pressure in the shorter tube is 1.0 mm of mercury, what is the pressure in the longer tube? 


ANSWER: According to Paschen's law, the sparking potential in discharge tubes is a function of the product of the pressure of the gas and the length of the tube. So,

Sparking potential V=f(pd) 

Since the sparking potential in the given case is the same for both tubes, hence products of pressure and length in both tubes must be equal. Thus, 

p₁d₁ =p₂d₂

→p₂ =p₁d₁/d₂

    =(1.0 mm)*(10 cm)/(20 cm)

   =0.50 mm of mercury.

         




 

     5.  Calculate n(T)/n(1000 K) for tungsten emitter at T = 300 K, 2000 K, and 3000 K where n(T) represents the number of thermions emitted per second by the surface at temperature T. Work function of tungsten is 4.52 eV. 


ANSWER: From the Richardson-Dushman equation, the thermionic current is given as

i =n(T)*e' =AST²e-φ/kT 

where e' is the charge on an electron and 'e' is the exponential e. S is the surface area, A is a constant related to the nature of the metal, T is the absolute temperature of the surface, φ is the work function of the metal and k is the Boltzmann constant. Considering two temperatures of the metal surface T and T', the ratio of the number of thermions emitted, 

n(T)/n(T') =(T/T')²eφ/kT'-φ/kT

   =(T/T')²e(1/T'-1/T)φ/k

Here T' =1000 K, 

φ =4.52 eV =4.52*1.6x10⁻¹⁹ J

k =1.38x10⁻²³ J/K

Hence φ/k =5.24x10⁴ K

So,

n(T)/n(1000 K) =(T/1000)²e5.24x10⁴(1/1000 -1/T)


Thus,

n(300 K)/n(1000 K)=(0.3)²e5.24x10⁴*(-7/3000)

          =0.09*e^-122.3

          =6.91x10⁻⁵⁵ 

  

n(2000 K)/n(1000 K) =2²*e5.24x10⁴*0.0005

    =4*e26.2

    =9.56x10¹¹.


And for T =3000 K, 

n(3000 K)/n(1000 K) =3²*e5.24x10⁴(2/3000)

 =9*e34.93

  =1.33x10¹⁶            






 

     6.  The saturation current from a thoriated-tungsten cathode at 2000 K is 100 mA. What will be the saturation current for a pure tungsten cathode of the same surface area operating at the same temperature? The constant A in the Richardson-Dushman equation is 60x10⁴ A/m²-K² for pure tungsten and 3.0x10⁴ A/m²-K² for thoriated-tungsten. The work function of pure tungsten is 4.5 eV and that of the thoriated tungsten is 2.6 eV. 


ANSWER: From the Richardson-Dushman equation, saturation current,

i =AST²e-φ/kT 

In the case of Thoriated Tungsten Cathode, i =100 mA =0.1 A, T =2000 K, A =3.0x10⁴ A/m²-K². Hence,

0.1 =3.0x10⁴S*(2000)²e-2.6x1.6x10⁻¹⁹/kT --(i)


For the pure Tungsten Cathode,

 A=60x10⁴ A/m²-K², φ =4.5 eV. So the saturation current,

i =60x10⁴S*(2000)²e-4.5x1.6x10⁻¹⁹/kT  -- (ii)


Dividing (ii) by (i),

i/0.1 =20*e(-1.9x1.6x10⁻¹⁹)/kT

{Here, k =1.38x10⁻²³ J/K, T =2000 K}

→i =2*e-11.01 A =3.3x10⁻⁵ A

    =33x10⁻⁶ A =33 µA.

     




 

     7.  A tungsten cathode and a thoriated-tungsten cathode have the same geometrical dimensions and are operated at the same temperature. The thoriated-tungsten cathode gives 5000 times more current than the other one. Find the operating temperature. Take relevant data from the previous problem. 


ANSWER: For the Tungsten Cathode, current

i =AST²e-φ/kT 

For the Thoriated Tungsten cathode, current

i' =A'ST²e-φ'/kT 

But given that, i' =5000i. So by taking the ratio, 

i'/i =(A'/A)e(φ-φ')/kT

→5000 =(3/60)e(4.5-2.6)*1.6x10⁻¹⁹/kT

→e1.9*1.6x10⁻¹⁹/kT =20*5000 =10⁵


Taking the log of both sides, we get

1.9*1.6x10⁻¹⁹/kT =5*ln (10) =11.513 

→kT =3.04x10⁻¹⁹/11.513

→T =3.04x10⁻¹⁹/(11.513*1.38x10⁻²³)

    =1913.4 K.            





 

     8.  If the temperature of the tungsten filament is raised from 2000 K to 2010 K, by what factor does the emission current change? The work function of tungsten is 4.5 eV. 


ANSWER: T =2000 K, T' =2010 K. 

Emission current for T', 

i' =AST'²e-φ/kT', 

and for T,

i =AST²e-φ/kT.

Hence the factor by which the emission current change is,

i'/i =(T'/T)²e(1/T -1/T')φ/k

  =(2010/2000)²e(1/2000-1/2010)4.5*1.6x10⁻¹⁹/1.38x10⁻²³

  =1.01*e0.13 

  =1.15.       






 

     9.  The constant A in the Richardson-Dushman equation for tungsten is 60x10⁴ A/m²-K². The work function of tungsten is 4.5 eV. A tungsten cathode having a surface area of 2.0x10⁻⁵ m² is heated by a 24 W electric heater. In a steady state, the heat radiated by the cathode equals the energy input by the heater and the temperature becomes constant. Assuming that the cathode radiates like a black body, calculate the saturation current due to thermions. Take Stefen constant =6x10⁻⁸ W/m²-K⁴. Assume that the thermions take only a small fraction of the heat supplied. 


ANSWER: The energy given to the cathode by the heater is radiated fully in a steady state. Since the cathode radiates like a black body, its emissivity is 1. From the Stefens law of black body radiation, power radiated by a black body is, 

P =𝞂AT⁴

→T⁴ =P/𝞂A

  =24/(6x10⁻⁸*2.0x10⁻⁵)

  =2x10¹³ =20x10¹²

→T =√(20 *10³

     =2114.7 K

Saturation current due to thermions,

i =AST²e-φ/kT

=60x10⁴*2x10⁻⁵*(2114.7)²*e-4.5*1.6x10⁻¹⁹/(1.38x10⁻²³*2114.7) A

=12*4.472x10⁶*e-24.672 A

=0.001 A

=1.0 mA. 






 

     10.  A plate current of 10 mA is obtained when 60 volts are applied across the diode tube. Assuming the Langmuir-Child equation iₚ ∝ Vₚ3/2 to hold, find the dynamic resistance rₚ in this operating condition. 


ANSWER: The dynamic resistance is given as,

rₚ =dVₚ/diₚ.

From the Langmuir-Child equation,

iₚ =kVₚ3/2,    ----------- (i)

where k is the constant of proportionality. Differentiating both sides,

diₚ =k*(3/2)Vₚ1/2 *dVₚ  ------(ii) 

Dividing (ii) by (i), 

diₚ/iₚ =(3/2)Vₚ⁻¹ *dVₚ

→dVₚ/diₚ =(2/3)Vₚ/iₚ

→rₚ =(2/3)*60/(10x10⁻³) Ω

      =4000 Ω 

     =4 kΩ.   

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Links to the Chapters




CHAPTER- 41- Electric Current Through Gases



CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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