JEE Main 2019 (Online) 10th April Evening Slot | Dual Nature of Radiation Question 120 | Physics

A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is : [Given Planck's constant h = 6.6 × 10^–34 Js, speed of light c = 3.0 × 10⁸ m/s]

5 × 10¹⁵

1.5 × 10¹⁶

1 × 10¹⁶

2 × 10¹⁶

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelentgh of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be:
Given E (in eV) = 1237/$$\lambda $$ (in nm)

4.5 eV

15.1 eV

3.0 eV

1.5 eV

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths '$$\lambda $$_x' and '$$\lambda $$_y' respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is :-

$${\lambda _x} - {\lambda _y}$$

$${{{\lambda _x}{\lambda _y}} \over {\left| {{\lambda _x} - {\lambda _y}} \right|}}$$

$${\lambda _x} + {\lambda _y}$$

$${{{\lambda _x}{\lambda _y}} \over {{\lambda _x} + {\lambda _y}}}$$

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

The electric field of light wave is given as $$$\overrightarrow E = {10^{ - 3}}\cos \left( {{{2\pi x} \over {5 \times {{10}^{ - 7}}}} - 2\pi \times 6 \times {{10}^{14}}t} \right)\mathop x\limits^ \wedge {{\rm N} \over C}$$$ This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is :
Given, E (in eV) = 12375/$$\lambda $$(inÅ)

2.48 V

0.48 V

0.72 V

2.0 V

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