1
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
2
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
$${{{\lambda _x}{\lambda _y}} \over {{\lambda _x} + {\lambda _y}}}$$
3
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
4
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
A nucleus A, with a finite de-broglie wavelength $$\lambda$$A, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths $$\lambda$$B and $$\lambda$$C of B and C are respectively :
A
$$\lambda$$A, 2$$\lambda$$A
B
2$$\lambda$$A, $$\lambda$$A
C
$$\lambda$$A, $$\lambda$$A/2
D
$$\lambda$$A/2, $$\lambda$$A
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