1
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$${{{\lambda _x}{\lambda _y}} \over {{\lambda _x} + {\lambda _y}}}$$
2
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
3
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
4
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two particles move at right angle to each other. Their de-Broglie wavelengths are $$\lambda _1$$ and $$\lambda _2$$ respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength $$\lambda _2$$ of the final particle, is given by :
A
$$\lambda = {{{\lambda _1} + {\lambda _2}} \over 2}$$
B
$${1 \over {{\lambda ^2}}} = {1 \over {\lambda _1^2}} + {1 \over {\lambda _2^2}}$$
C
$$\lambda = \sqrt {{\lambda _1}{\lambda _2}} $$
D
$${2 \over \lambda } = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$
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