NEET UG PHYSICS MCQ: SEMI CONDUCTOR DEVICES

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The acceptor level of a p-type semiconductor is

6 \mathrm{~eV}

. The maximum wavelength of light which can create a hole would be : Given

\mathrm{hc}=1242 \mathrm{~eV} \mathrm{~nm}

View answer and explanation

Correct option: C

207 nm

Option C (207 nm) is correct because the maximum wavelength corresponds to the minimum energy required to create a hole, which is equal to the acceptor level energy. Using the formula

E = \frac{hc}{\lambda}

, we can solve for

\lambda

More Questions from SEMI CONDUCTOR DEVICES

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Following is the relation between current and charge

I = A{T^2}{e^{q/{V_L}}}

, then value of

{V_L}

will be To measure light intensity we use The ratio of RMS output voltage of full wave rectifier and half wave rectifier is

\sqrt{n}.

The value of

n

is? (The input voltage is sine wave). The truth table shown in figure is for
A    0   0   1   1
B    0   1   0   1
Y    1   0   0   1  The average value of output direct current in a half wave rectifier is A

p - n\;

junction has acceptor impurity concentration of

{10^{17}}\mathrm{c}{\mathrm{m}^{ - 3}}

in the

P

side and donor impurity concentration of

{10^{16}}

\mathrm{c}{\mathrm{m}^{ - 3}}

in the

N

side. What is the contact potential at the junction? (

kT =

thermal energy, intrinsic carrier concentration

{n_i} = 1.4 \times {10^{10}}

\mathrm{c}{\mathrm{m}^{ - 3}}

)