A potential difference of 2 V is applied between the opposite faces of a Ge crystal plate of area 1 cm 2 and thickness 0-5 mm- If the concentration of electrons in Ge is 2 - 1019 - m 3 and mobilities of electrons and holes are 0-36 m 2- volt -sec and 0-14 m 2- volt -sec respectively- then the current flowing through the plate will be A- 0-25 AB- 0-45 AC- 0-56 AD- 0-64 A

The correct option is D 0.64 AConductivity σ = ne(μ_e + μ_h)=2 × 10^19× 1.6 × 10^ 19 (0.36 +0.14) σ = 1.6 (Ω m)^ 1 Resistance R = ρ lA = lσ A R = 0.5 × 10^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Solenoid

A potential d...

Question

A potential difference of $2 V$ is applied between the opposite faces of a Ge crystal plate of area $1 {cm}^{2}$ and thickness $0.5 mm$ . If the concentration of electrons in Ge is $2 \times 10^{19} / m^{3}$ and mobilities of electrons and holes are $0.36 \frac{m^{2}}{volt -sec}$ and $0.14 \frac{m^{2}}{volt -sec}$ respectively, then the current flowing through the plate will be

0.25 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.45 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.56 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.64 A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $0.64 A$
Conductivity $σ = n e (μ_{e} + μ_{h}) = 2 \times 10^{19} \times 1.6 \times 10^{- 19} (0.36 + 0.14)$
$σ = 1.6 (Ω - m)^{- 1}$
Resistance $R = \frac{ρ l}{A} = \frac{l}{σ A}$
$R = \frac{0.5 \times 10^{- 3}}{1.6 \times 10^{- 4}} = \frac{25}{8} Ω$
From Ohms law
$i = \frac{V}{R} = \frac{2}{\frac{25}{8}} = \frac{16}{25}$
$i = 0.64 A$

Suggest Corrections

Similar questions

Q. A potential difference of

2 V

is applied across the opposite faces of a Germanium plate, of area

1 {cm}^{2}

and thickness

0.5 mm

. If the concentration of electrons is

2 \times 10^{19} m^{- 3}

and mobilities of electrons and holes are

0.36 m^{2} /V-s

and

0.14 m^{2} /V-s

respectively, the current flowing through the plate will be,

Q. A potential difference of

2.5

V is applied across the faces of a germanium crystal plate. The face area of the crystal is

1

c m^{2}

and its thickness is

1.0

mm. The free electron concentration in germanium is

2 \times 10^{19} m^{- 3}

and the electron and holes mobilities are

0.33 m^{2} / V

and

0.17 m^{2} / V

respectively. The current across the plate will be?

Q. In a semiconductor, the concentrations of electrons and holes are

8 \times 10^{18} / m^{3}

and

5 \times 10^{18} / m^{3}

respectively. If the mobilities of electrons and hole are

2.3 m^{2}

/volt-sec and

0.01 m^{2}

/volt-sec respectively, then semiconductor is:

Mobilities of electrons and holes in a sample of intrinsic germanium semiconductor
at room temperature are 0.36m²/volt-sec and 0.17 m²/volt-sec respectively. If the electron and hole densities are each equal to 2.5 X 10¹⁹/m³, the conductivity is.

Q. Mobilities of electrons and holes in a sample of intrinsic Ge at room temperature are

0.35 m^{2} / V - s

and

0.18 m^{2} / V - s

respectively. If the electron and hole densities are each equal to

2.5 \times 10^{19} / m^{3}

, the Ge conductivity will be

Join BYJU'S Learning Program

The correct option is D 0.64 AConductivity σ=ne(μe+μh)=2×1019×1.6×10−19(0.36+0.14) σ=1.6 (Ω−m)−1 Resistance R=ρlA=lσA R=0.5×10−31.6×10−4=258Ω From Ohms law i=VR=2258=1625 i=0.64 A