1
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
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]
A
5 × 1015
B
1.5 × 1016
C
1 × 1016
D
2 × 1016
2
JEE Main 2019 (Online) 10th April Morning Slot
+4
-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)
A
4.5 eV
B
15.1 eV
C
3.0 eV
D
1.5 eV
3
JEE Main 2019 (Online) 9th April Evening Slot
+4
-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 :-
A
$${\lambda _x} - {\lambda _y}$$
B
$${{{\lambda _x}{\lambda _y}} \over {\left| {{\lambda _x} - {\lambda _y}} \right|}}$$
C
$${\lambda _x} + {\lambda _y}$$
D
$${{{\lambda _x}{\lambda _y}} \over {{\lambda _x} + {\lambda _y}}}$$
4
JEE Main 2019 (Online) 9th April Morning Slot
+4
-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)
A
2.48 V
B
0.48 V
C
0.72 V
D
2.0 V
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