Semiconductors and Semiconductor Devices

Concept Of Physics

H C Verma

1   Calculate the number of states per cubic metre of sodium in 3s band. The density of sodium is 1013 kg/m . How many of them are empty ?

Solution :

$\\$ f= 1013 $\frac{kg}{m^3}, V=1m^3$ $\\$ m= fV= 1013 $\times 1$ = 1013 kg $\\$ No. of atoms = $\frac{1013 \times 10^3 \times 6 \times 10^{23}}{23} = 264.26 \times 10^{26}.$ $\\$ a) Total no.of states = $2N = 2 \times 264.26 \times 10 = 528.52 = 5.3 \times 10^{28} \times 10^{26}$ $\\$ b) Total no.of unoccupied states = 2.65 $\times 10^{26}.$

2   In a pure semiconductor, the number of conduction electrons is $6 \times 10^{19}$ per cubic metre. How many holes are there in a sample of size $1 cm \times 1 cm \times 1 mm$ ?

Solution :

$\\$ In a pure semiconductor, the no.of conduction electrons = no.of holes $\\$ Given volume = $1 cm \times 1 cm \times 1 mm$ $\\$ $= 1 \times 10^{-2} \times 1 \times 10^{-2} \times 1 \times 10^{-3} = 10^{-7} m^3$ $\\$ No. of electrons = $6 \times 10^{19} \times 10^{-7} = 6 \times 10^{12}.$ $\\$ Hence no.of holes = $6 \times 10^{12}.$

3   Indium antimonide has a band gap of 0.23 eV between the valence and the conduction band. Find the temperature at which kT equals the band gap.

Solution :

$\\$ E = 0.23 eV, K = 1.38 $\times 10^{-23}$ $\\$ KT = E $\\$ $\Rightarrow 1.38 \times 10^{-23} \times T = 0.23 \times 1.6 \times 10^{-19}$ $\\$ $\Rightarrow T = \frac{0.23 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23}} = \frac{0.23 \times 1.6 \times 10^4}{1.38} = 0.2676 \times 10^4 = 2670.$

4   The band gap for silicon is 11 eV. (a) Find the ratio of the band gap to kT for silicon at room temperature 300 K. (b) At what temperature does this ratio become one tenth of the value at 300 K ? (Silicon will not retain its structure at these high temperatures.)

Solution :

$\\$ Bandgap = 1.1 eV, T = 300 K $\\$ a) Ratio = $\frac{1.1}{KT} = \frac{1.1}{8.62 \times 10^{-5} \times 3 \times 10^2} = 42.53 = 43$ $\\$ b) 4.253' = $\frac{1.1}{8.62 \times 10^{-5} \times T} or T = \frac{1.1 \times 10^5}{4.253 \times 8.62} = 3000.47 K$

5   When a semiconducting material is doped with an impurity, new acceptor levels are created. In a particular thermal collision, a valence electron receives an energy equal to 2kT and just reaches one of the acceptor levels. Assuming that the energy of the electron was at the top edge of the valence band and that the temperature T is equal to 300 K, find the energy of the acceptor levels above the valence band.

Solution :

$\\$ 2KT = Energy gap between acceptor band and valency band $\\$ $\Rightarrow 2 \times 1.38 \times 10^{-23} \times 300$ $\Rightarrow E = (2 \times 1.38 \times 3) \times 10^{-21}J = \frac{6 \times 1.38}{1.6} \times \frac{10^{-21}}{10^{-19}}eV = (\frac{6\times 1.38}{1.6}) \times 10^{-2}eV$ $\\$ $= 5.175 \times 10^{-2}eV = 51.75meV = 50 meV.$

6   The band gap between the valence and the conduction bands in zinc oxide (ZnO) is 3'2 eV. Suppose an electron in the conduction band combines with a hole in the valence band and the excess energy is released in the form of electromagnetic radiation. Find the maximum wavelength that can be emitted in this process.

Solution :

$\\$Given : Band gap = 3.2 eV, $\\$ $E = \frac{hc}{\lambda} = \frac{1242}{\lambda} = 3.2$ or $\lambda = 1.9 \times 10^{-5}m.$

7   Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820 nm, what is the band gap ?

Solution :

$\\$ $\lambda = 820 nm$ $\\$ $E= \frac{hc}{\lambda} = \frac{1242}{820} = 1.5 eV$

8   Find the maximum wavelength of electromagnetic radiation which can create a hole--electron pair in germanium. The band gap in germanium is 0'65 eV.

Solution :

$\\$ Band Gap = 0.65 eV, $\lambda = ?$ $\\$ $E = \frac{hc}{\lambda} = \frac{1242}{0.65} = 1910.7 \times 10^{-9}m = 1.9 \times 10^{-5}m.$

9   In a photodiode, the conductivity increases when the material is exposed to light. It is found that the conductivity changes only if the wavelength is less than 620 nm. What is the band gap ?

Solution :

$\\$ Band gap = Energy need to over come the gap $\\$ $\frac{hc}{\lambda} = \frac{1242eV-nm}{620nm} = 2.0eV.$

10   Let $\Delta$E denote the energy gap between the valence band and the conduction band. The population of conduction electrons (and of the holes) is roughly proportional to $e^{\frac{\Delta E_1}{2KT}}$ Find the ratio of the concentration of conduction electrons in diamond to that in silicon at room temperature 300 K. AE for silicon is 1.1 eV and for diamond is 6 0 eV. How many conduction electrons are likely to be in one cubic metre of diamond ?

Solution :

$\\$ Given n = $e^{\frac{\Delta E}{2KT}}, \Delta E = Diamon \rightarrow 6 eV ; \Delta E Si \rightarrow 1.1 eV$ $\\$ Now, $n_1 = e^{-\frac{\Delta E_1}{2KT}} = e^{\frac{-6}{2 \times 300 \times 8.62 \times 10^{-5}}}$ $\\$ $n_2 = e^{-\frac{\Delta E_2}{2KT}} = e^{\frac{-1.1}{2 \times 300 \times 8.62 \times 10^{-5}}}$ $\\$ $\frac{n_1}{n_2} = \frac{4.14772 \times 10^{-51}}{5.7978 \times 10^{-10}} = 7.15 \times 10^{-42}.$ $\\$ Due to more $\Delta E$, the conduction electrons per cubic metre in diamond is almost zero.

11   The conductivity of a pure semiconductor is roughly proportional to $T ^{\frac{3}{2}}$ $e^{\frac{\Delta E}{2KT}}$ where AE is the band gap. The band gap for germanium is 0.74 eV at 4 K and 0.67 eV at 300 K. By what factor does the conductivity of pure germinium increase as the temperature is raised from 4 K to 300 K ?

Solution :

$\\$ $\sigma = T^{\frac{3}{2}} e^{-\frac{\Delta E_1}{2KT}} at 4^o K$ $\\$ $\sigma = 4^{\frac{3}{2}} = e^{\frac{-0.74}{2\times 8.62 \times 10^{-5} \times 4}} = 8 \times e^{-1073.08}.$ $\\$ At 300 K, $\\$ $\sigma = 300^{\frac{3}{2}} e^{\frac{-0.67}{2\times 8.62 \times 10^{-5} \times 300}} = \frac{3 \times 1730}{8} e^{-12.95}.$ $\\$ Ratio = $\frac{8\times e^{1073.08}}{[\frac{(3\times 1730)}{8}\times e^{-12.95}]} = \frac{64}{3\times 1730} e^{-1060.13}.$

12   Estimate the proportion of boron impurity which will increase the conductivity of a pure silicon sample by a factor of 100. Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same temperature is $7 \times 10^{15}$ holes per cubic metre. Density of silicon is $5 \times 10^{15}$ atoms per cubic metre.

Solution :

$\\$ Total no.of charge carriers initially = $2 \times 7 \times 10^{15} = 14 \times 10^{15}$/ Cubic meter $\\$ Finally the total no.of charge carriers = $14 \times \frac{10^{17}}{m^3}$ $\\$ We know : The product of the concentrations of holes and conduction electrons remains, almost the same. Let x be the no.of holes. $\\$ So, $(7\times 10^{15}) \times (7\times 10^{15}) = x \times (14 \times 10^{17} - x)$ $\\$ $\Rightarrow 14x \times 10^{17} - x^2 = 79 \times 10^{30}$ $\\$ $\Rightarrow x^2 - 14x \times 10^{17} - 49 \times 10^{30} = 0$ $\\$ $x= \frac{14 \times 10^{17} \pm 14^2 \times \sqrt{10^{34}+4\times 49 \times 10^{30}}}{2} = 14.00035 \times 10^{17}.$ $\\$ = Increased in no.of holes or the no.of atoms of Boron added. $\\$ $\Rightarrow$ 1 atom of Boron is added per $\frac{5\times 10^{28}}{1386.035 \times 10^{15}} = 3.607 \times 10^{-3} \times 10^{13} = 3.607 \times 10^{10}.$

13   The product of the hole concentration and the conduction electron concentration turns out to be independent of the amount of any impurity doped. The concentration of conduction electrons in germanium is $6 \times 10^{19}$ per cubic metre. When some phosphorus impurity is doped into a germanium sample, the concentration of conduction electrons increases to $2 \times 10^{23}$ per cubic metre. Find the concentration of the holes in the doped germanium.

Solution :

$\\$ (No. of holes) (No.of conduction electrons) = constant. $\\$ At first : No. of conduction electrons = $6 \times 10^{19}$ No.of holes = $6 \times 10^{19}$ $\\$ After doping No.of conduction electrons = $2 \times 10^{23}$ No. of holes = x. $\\$ $(6 \times 10^{19}) (6 \times 10^{19}) = (2 \times 10^{23})x$ $\\$ $\Rightarrow \frac{6\times 6\times 10^{19+19}}{2\times 10^{23}} = x$ $\\$ $\Rightarrow x = 18 \times 10^{15} = 1.8\times 10^{16}.$

14   The conductivity of an intrinsic semiconductor depends on temperature as$\sigma = \sigma_n e^{\frac{\Delta E}{2KT}}$ where a„ is a constant. Find the temperature at which the conductivity of an intrinsic germanium semiconductor will be double of its value at T ~ 300 K. Assume that the gap for germanium is 0.650 eV and rtemains constant as the temperature is increased.

Solution :

$\\$ $\sigma = \sigma_0 e^{\frac{\Delta E}{2KT}}$ $\\$ $\Delta E = 0.650 eV, T=300 K$ $\\$ According to question, K = 8.62 $\times 10^{-5} eV$ $\\$ $\sigma_0 e^{\frac{\Delta E}{2KT}} = 2 \times \sigma_0 e^{\frac{-\Delta E}{2\times K \times 300}}$ $\\$ $\Rightarrow e^{\frac{-0.65}{2\times 8.62 \times 10^{-5} \times T}} 6.96561 \times 10^{-5}$ $\\$ Taking in on both sides, $\\$ We get, $\frac{-0.65}{2 \times 8.62 \times 10^{-5} \times T'} = -11.874525$ $\\$ $\Rightarrow \frac{1}{T'} = \frac{11.574525 \times 2 \times 8.62 \times 10^{-5}}{0.65}$ $\\$ $\Rightarrow T' = 317.51178 = 318K.$

15   A semiconducting material has a band gap of 1 eV. Acceptor impurities are doped into it which create acceptor levels 1 meV above the valence band. Assume that the transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap. Also, if kT is more than twice the gap, the upper levels have maximum population. The temperature of the semiconductor is increased from 0 K. The concentration of the holes increases with temperature and after a certain temperature it becomes approximately constant. As the temperature is further increased, the hole concentration again starts increasing at a certain temperature. Find the order of the temperature range in which the hole concentration remains approximately constant.

Solution :

$\\$ Given band gap = 1 eV $\\$ Net band gap after doping = $(1-10^{-3})eV = 0.999eV$ $\\$ According to the question, $KT_1 = 0.999/50$ $\\$ $\Rightarrow T_1 = 231.78 = 231.8$ $\\$For the maximum limit $KT_2 = 2 \times 0.999$ $\\$ $\Rightarrow T_2 = \frac{2\times 1 \times 10^{-3}}{8.62\times 10^{-5}} = \frac{2}{8.62} \times 10^2 = 23.2.$ $\\$ Temperature range is (23.2 – 231.8).

16   In a p-n junction, the depletion region is 400 nm wide and an electric field of $5 \times 10^5$ V/m exists in it. (a) Find the height of the potential barrier, (b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side ?

Solution :

$\\$ Depletion region ‘d’= 400 nm = $4 \times 10^{-7}m$ $\\$ Electric field E = 5 $\times 10^{5} \frac{V}{m}$ $\\$ a) Potential barrier V = $E \times d = 0.2 V$ $\\$ b) Kinetic energy required = Potential barrier $\times$ e = 0.2 eV [Where e = Charge of electron]

17   The potential barrier existing across an unbiased p-n junction is 0'2 volt. What minimum kinetic energy a hole should have to diffuse from the p-side to the n-side if (a) the junction is unbiased, (b) the junction is forward- biased at O'l volt and (c) the junction is reverse-biased at O'l volt?

Solution :

$\\$ Potential barrier = 0.2 Volt $\\$ a) K.E. = (Potential difference) $\times$ e = 0.2 eV (in unbiased $cond^n$) $\\$ b) In forward biasing $\\$ KE + Ve = 0.2e $\\$ $\Rightarrow$ KE = 0.2e – 0.1e = 0.1e. $\\$ c) In reverse biasing $\\$ KE – Ve = 0.2 e $\\$ $\Rightarrow$ KE = 0.2e + 0.1e = 0.3e.

18   In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction (a) from the p-side and (b) from the /i-side.

Solution :

$\\$ Potential barrier ‘d’ = 250 meV $\\$ Initial KE of hole = 300 meV $\\$ We know : KE of the hole decreases when the junction is forward biased and increases when reverse blased in the given ‘Pn’ diode. So, $\\$a) Final KE = (300 – 250) meV = 50 meV $\\$ b) Initial KE = (300 + 250) meV = 550 meV

19   When a p-n junction is reverse-biased, the current becomes almost constant at 25 fjA. When it is forward- biased at 200 mV, a current of 75 pA is obtained. Find the magnitude of diffusion current when the diode is (a) unbiased, (b) reverse-biased at 200 mV and (c) forward-biased at 200 mV

Solution :

$\\$ $i_1 = 25\mu A, V = 200 mV, i_2 = 75 \mu A$ $\\$ a) When in unbiased condition drift current = diffusion current $\\$ $\therefore Diffusion current = 25 \mu A.$ $\\$ b) On reverse biasing the diffusion current becomes ‘O’. $\\$ c) On forward biasing the actual current be x. $\\$ x – Drift current = Forward biasing current. $\\$ $\Rightarrow x – 25 \mu A = 75 \mu A$ $\\$ $\Rightarrow x = (75 + 25) \mu A = 100 \mu A.$

20   The drift current in a p-n junction is 20 0 pA. Estimate the number of electrons crossing a cross-section per second in the depletion region.

Solution :

$\\$ Drift current = $20 \mu A = 20 \times 10^{-6} A.$ $\\$ Both holes and electrons are moving. $\\$ So, no.of electrons = $\frac{20 \times 10^{-6}}{2\times 1.6 \times 10^{-19}} = 6.25 \times 10^{13}.$

21   The current-voltage characteristic of an ideal p-n junction diode is given by $i = i_0(e^{\frac{eV}{KT}})$ where the drift current i a equals 10 fiA. Take the temperature T to be 300 K. (a) Find the voltage V for which $\frac{eV}{KT}$ = 100. One can neglect the term 1 for voltages greater than this value, (b) Find an expression for the dynamic resistance of the diode as a function of V for V > V 0 . (c) Find the voltage for which the dynamic resistance is 0"2 SI.

Solution :

$\\$ a) $e^{\frac{aV}{KT}}$ = 100 $\\$ $\Rightarrow e^{\frac{V}{8.62\times 10^{-5} \times 300}}$ = 100 $\\$ $\Rightarrow e^{\frac{V}{8.62\times 10^{-5} \times 300}}$ = 4.605 $\Rightarrow V= 4.605 \times 8.62 \times 3 \times 10^{-3} = 119.08 \times 10^{-3}$ $R= \frac{V}{I} = \frac{V}{I_0(e^{\frac{ev}{KT}})} = \frac{119.08 \times 10^{-3}}{10 \times 10^{-6} \times (100-1)} = \frac{119.08 \times 10^{-3}}{99\times 10^{-5}} = 1.2 \times 10^{2}.$ $\\$ $V_0 = I_0R$ $\\$ $\Rightarrow 10 \times 10^{-6} \times 1.2 \times 10^2 = 1.2 \times 10^{-3} = 0.0012V.$ $\\$c)0.2 = $\frac{KT}{ei_0}e^{\frac{eV}{KT}}$ $\\$ k=8.62 $\times 10^{-5} \frac{eV}{k}, t=300 K$ $\\$ $i_0 = 10 \times 10^{-5} A.$ $\\$ Substituting the values in the equation and solving $\\$ We get V = 0.25

22   Consider a p-n junction diode having the characteristic $i = i _0(e^{\frac{aV}{KT}} - 1)$ where i 0 = 20 |uA. The diode is operated at T = 300 K. (a) Find the current through the diode when a voltage of 300 mV is applied across it in forward bias, (b) At what voltage does the current double ?

Solution :

$\\$ a) $i_0 = 20 \times 10^{-6}A$, T=300k, V=300mV $\\$ i= $i_0 e^{\frac{ev}{KT}-1} = 20 \times 10^{-6} (e^{\frac{100}{8.62} -1}) = 2.18 A = 2A$ b)4 = $20 \times 10^{-6} (e^{\frac{V}{8.62 \times 3 \times 10^{-2}} -1}) \Rightarrow e^{\frac{V \times 10^3}{8.62 \times 3} -1} = \frac{4 \times 10^6}{20}$ $\\$ $\Rightarrow e^{\frac{V \times 10^3}{8.62 \times 3}} = 200001 \Rightarrow \frac{V \times 10^3}{8.62 \times 3} = 12.2060$ $\\$ $\Rightarrow$ V = 315 mV = 318 mV.

23   Calculate the current through the circuit and the potential difference across the diode shown in figure (45-E1). The drift current for the diode is 20 $\mu A$.

Solution :

$\\$ a) Current in the circuit = Drift current (Since, the diode is reverse biased = 20 $\mu A$) $\\$ b) Voltage across the diode = 5-($20 \times 20 \times 10^{-6}$) $\\$ = $5-(4 \times 10^{-4}) = 5 V$.

24   Each of the resistances shown in figure (45-E2) has a value of 20 Q. Find the equivalent resistance between A and B. Does it depend on whether the point A or B is at higher potential ?

Solution :

$\\$ From the figure : According to wheat stone bridge principle, there is no current through the diode. $\\$ Hence net resistance of the circuit is $\frac{40}{2} = 20 \Omega$.

25   Find the-currents through the resistances in the circuits shown in figure (45-E3).

Solution :

$\\$ a) Since both the diodes are forward biased net resistance = 0 $\\$ $i= \frac{2V}{2 \Omega} = 1 A$ $\\$ b) One of the diodes is forward biased and other is reverse biase. Thus the resistance of one becomes $\infty$. $\\$ $i = \frac{2}{2 + \infty} = 0 A.$ $\\$ Both are forward biased. Thus the resistance is 0. $\\$ $i= \frac{2}{2} = 1A.$ $\\$ One is forward biased and other is reverse biased. Thus the current passes through the forward biased diode. $\\$ $\therefore i= \frac{2}{2} = 1A.$

26   What are the readings of the ammeters A, and A., shown in figure (45-E4). Neglect the resistances of the meters.

Solution :

$\\$ The diode is reverse biased. Hence the resistance is infinite. So, current through A 1 is zero. $\\$ $For A_2, current = \frac{2}{10} =m0.2Amp.$

27   Find the current through the battery in each of the circuits shown in figure (45-E5).

Solution :

$\\$Both diodes are forward biased. Thus the net diode resistance is 0. $\\$ $i= \frac{5}{(10+10)/10.10} = \frac{5}{5} =1A.$ One diode is forward biased and other is reverse biased. $\\$ Current passes through the forward biased diode only. $\\$ $i= \frac{V}{R_{net}} = \frac{5}{10+0} = \frac{1}{2} = 0.5A.$

28   Find the current through the resistance R in figure (45-E6) if (a) R = 12 $\Omega$. (b) R = 48$\Omega$.

Solution :

$\\$a) When R = 12 $\Omega$ The wire EF becomes ineffective due to the net (–)ve voltage. Hence, current through R = 10/24 = 0.4166 = 0.42 A. $\\$ b) Similarly for R = 48 $\Omega$. $\\$ $i= \frac{10}{48+12} = \frac{10}{60}= 0.16A.$

29   Draw the current-voltage characteristics for the device shown in figure (45-E7) between the terminals A and B.

Solution :

$\\$ Since the diode 2 is reverse biased no current will pass through it.

30   Find the equivalent resistance of the network shown in figure (45-E8) between the points A and B.

Solution :

$\\$ Let the potentials at A and B be $V_A$ and $V_B$ respectively. $\\$ i) If $V_A$ > $V_B$ Then current flows from A to B and the diode is in forward biased. Eq. Resistance = 10/2 = 5 $\Omega$. $\\$ ii) If $V_A$ < $V_B$ Then current flows from B to A and the diode is reverse biased. Hence Eq.Resistance = 10 $\Omega$.

31   When the base current in a transistor is changed from $30 \mu A$ to $80 \mu A$ , the collector current is changed from l'O M A to 3.5 M A . Find the current gain $\beta$.

Solution :

$\\$ $\beta = 50, \delta l_b = 50 \mu A,$ $\\$ $V_0 = \beta \times RG = 50 \times \frac{2}{0.5} = 200.$ $\\$ a) VG = $\frac{V_0}{v_1} = \frac{V_0}{V_i} = \frac{V_0}{\delta l_b \times R_i} = \frac{200}{50\times 10^{-6} \times 5 \times 10^{2}} = 8000V.$ $\\$ b) $\delta V_i = \delta l_b \times R_i = 50 \times 10^{-6} \times 5 \times 10^2 = 0.00025V = 25mV.$ $\\$c) Power gain = $\beta^2 \times RG = \beta^2 \times \frac{R_0}{R_i} = 2500 \times \frac{2}{0.5} = 10^4.$

32   Let X = ABC + BCA + CAB. Evaluate X for (a) A = 1, B = 0, C = 1, (b) A = B = C = 1, (c) A = B = C = 0.

Solution :

$\\$ X = $A\vec{BC} + B\vec{CA} + C\vec{AB}$ $\\$ a) A = 1, B = 0, C = 1 $\\$ X = 1. $\\$b) A = B = c = 1 $\\$ X = 0.

33   Design a logical circuit using AND, OR and NOT gates to evaluate ABC + BCA.

Solution :

Logical Circuit Diagram

34   Show that AB + AB is always 1.

Solution :

$\\$LHS = $AB \times \vec{AB} = X + \vec{X}$ [X = AB] $\\$ if X = 0, $\vec{X} = 1$ $\\$ if $\vec{X} = 0$, X = 1 $\\$ $\Rightarrow 1 + 0$ or 0 + 1 = 1 $\\$ $\Rightarrow$ RHS = 1 (Proved)

35   The guru of a yogi lives in a Himalyan cave, 1000 km away from the house of the yogi. The yogi claims that whenever he thinks about his guru, the guru immediately knows about it. Calculate the minimum possible time interval between the yogi thinking about the guru and the guru knowing about it.

Solution :

$\\$S = 1000 km = $10^6$ m $\\$ The process requires minimum possible time if the velocity is maximum. $\\$ We know that maximum velocity can be that of light i.e. = $3 \times 10^8 m/s$. $\\$ So, time = $\frac{Distance}{Speed} = \frac{10^6}{3 \times 10^8} = \frac{1}{300}s$.

36   A suitcase kept on a shop's rack is measured 50 cm x 25 cm x 10 cm by the shop's owner. A traveller takes this suitcase in a train moving with velocity 06c. If the suitcase is placed with its length along the train's velocity, find the dimensions measured by (a) the traveller and (b) a ground observer.

Solution :

$\\$l = 50 cm, b = 25 cm, h = 10 cm, v = 0.6 C $\\$ a) The observer in the train notices the same value of l, b, h because relativity are in due to difference in frames. $\\$ b) In 2 different frames, the component of length parallel to the velocity undergoes contraction but the perpendicular components remain the same. So length which is parallel to the x-axis changes and breadth and height remain the same. $\\$ e' = $e\sqrt{1-\frac{V^2}{C^2}} = 50\sqrt{1-\frac{(0.6)^2 C^2}{C^2}}$ $\\$ = $50\sqrt{1-0.36} = 50 \times 0.8 = 40 cm.$ $\\$ The lengths observed are $40 cm \times 25 cm \times 10 cm.$