1
JEE Main 2022 (Online) 28th July Morning Shift
+4
-1

The equation $$\lambda=\frac{1.227}{x} \mathrm{~nm}$$ can be used to find the de-Brogli wavelength of an electron. In this equation $$x$$ stands for :

Where

$$\mathrm{m}=$$ mass of electron

$$\mathrm{P}=$$ momentum of electron

$$\mathrm{K}=$$ Kinetic energy of electron

$$\mathrm{V}=$$ Accelerating potential in volts for electron

A
$$\sqrt{\mathrm{mK}}$$
B
$$\sqrt{\mathrm{P}}$$
C
$$\sqrt{\mathrm{K}}$$
D
$$\sqrt{\mathrm{V}}$$
2
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1

With reference to the observations in photo-electric effect, identify the correct statements from below :

(A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light.

(B) The value of saturation current increases on moving the source of light away from the metal surface.

(C) The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light.

(D) The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves.

(E) Existence of threshold wavelength can not be explained by wave nature of light/ electromagnetic waves.

Choose the correct answer from the options given below :

A
(A) and (B) only
B
(A) and (E) only
C
(C) and (E) only
D
(D) and (E) only
3
JEE Main 2022 (Online) 27th July Morning Shift
+4
-1

An electron (mass $$\mathrm{m}$$) with an initial velocity $$\vec{v}=v_{0} \hat{i}\left(v_{0}>0\right)$$ is moving in an electric field $$\vec{E}=-E_{0} \hat{i}\left(E_{0}>0\right)$$ where $$E_{0}$$ is constant. If at $$\mathrm{t}=0$$ de Broglie wavelength is $$\lambda_{0}=\frac{h}{m v_{0}}$$, then its de Broglie wavelength after time t is given by

A
$$\lambda_{0}$$
B
$$\lambda_{0}\left(1+\frac{e E_{0} t}{m v_{0}}\right)$$
C
$$\lambda_{0} t$$
D
$$\frac{\lambda_{0}}{\left(1+\frac{e E_{0} t}{m v_{0}}\right)}$$
4
JEE Main 2022 (Online) 26th July Morning Shift
+4
-1

A parallel beam of light of wavelength $$900 \mathrm{~nm}$$ and intensity $$100 \,\mathrm{Wm}^{-2}$$ is incident on a surface perpendicular to the beam. The number of photons crossing $$1 \mathrm{~cm}^{2}$$ area perpendicular to the beam in one second is :

A
$$3 \times 10^{16}$$
B
$$4.5 \times 10^{16}$$
C
$$4.5 \times 10^{17}$$
D
$$4.5 \times 10^{20}$$
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