Tag: band theory of solids, a brief introduction

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Multiple choice physics semiconductors band theory of solids, a brief introduction electron energies in solids energy bands

The electrical conductivity of a semiconductor increases when radiation of the wavelength shorter than $2480nm$ is incident on it.The bandgap (in eV) for the semiconductor is

  1. $0.5$

  2. $0.9$

  3. $0.7$

  4. $1.1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that $E=\dfrac{hc}{\lambda}$

$\Rightarrow E=\dfrac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2480 \times 10^{-9}}=0.5 ev$
Therefore the band gap for the semiconductor is $0.5ev$
So the correct option is $A$

Multiple choice physics semiconductors band theory of solids, a brief introduction electron energies in solids energy bands

State whether given statement is True or False
The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature decreases exponentially with increasing bandgap.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature decreases exponentially with increasing bandgap because it is more difficult for the electrons to jump to the conduction band from valence band if the band gap between them is large.

Therefore the given statement is TRUE.

Multiple choice physics semiconductors band theory of solids, a brief introduction electron energies in solids energy bands

A transformer has 500 turns in its primary and 1000 turns in its secondary winding.The primary voltage is 200 V and the load in the secondary is 100 ohm.Calculate the current in the primary,assuming it to be a ideal transformer.

  1. 25 A

  2. 45 A

  3. 8A

  4. 22 A
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In usual rotation we have -

$\Rightarrow \dfrac{V _s}{V _p}=\dfrac{N _s}{N _p}$
$\therefore V _S=V _p\times \dfrac{N _s}{N _p}$
          $=200v\times \dfrac{1000}{500}$
          $=400V$
There is load resistance of $100\Omega$ in the secondary circuit. Therefore current in the secondary 
$\Rightarrow i _s=\dfrac{V _s}{R _s}=\dfrac{400V}{100\Omega}=4A$
In an ideal transformer the output power and input power are equal that is 
$\Rightarrow V _s\times i _s=V _p\times i _p$
$\therefore i _p=i _s \times \dfrac{V _s}{V _p}=4A\times \dfrac{400V}{200V}$
                            $=4A\times 2$
                            $=8A$
Hence, the answer is $8A.$