JEE Main 2024 (Online) 6th April Evening Shift | Semiconductor Question 6 | Physics

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

407 nm

103.5 nm

414 nm

207 nm

JEE Main 2024 (Online) 6th April Morning Shift

MCQ (Single Correct Answer)

-1

The correct truth table for the following logic circuit is :

JEE Main 2024 (Online) 6th April Morning Shift Physics - Semiconductor Question 1 English

JEE Main 2024 (Online) 6th April Morning Shift Physics - Semiconductor Question 1 English Option 3

JEE Main 2024 (Online) 6th April Morning Shift Physics - Semiconductor Question 1 English Option 4

JEE Main 2024 (Online) 5th April Evening Shift

MCQ (Single Correct Answer)

-1

The output (Y) of logic circuit given below is 0 only when :

JEE Main 2024 (Online) 5th April Evening Shift Physics - Semiconductor Question 3 English

$$\mathrm{A}=0, \mathrm{~B}=0$$

$$\mathrm{A}=0, \mathrm{~B}=1$$

$$\mathrm{A}=1, \mathrm{~B}=0$$

$$\mathrm{A=1, B=1}$$

JEE Main 2024 (Online) 5th April Morning Shift

MCQ (Single Correct Answer)

-1

Following gates section is connected in a complete suitable circuit.

JEE Main 2024 (Online) 5th April Morning Shift Physics - Semiconductor Question 4 English

For which of the following combination, bulb will glow (ON) :

$$\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=1, \mathrm{D}=0$$

$$\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0, \mathrm{D}=0$$

$$\mathrm{A}=0, \mathrm{~B}=0, \mathrm{C}=0, \mathrm{D}=1$$

$$\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=1, \mathrm{D}=1$$

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