1
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+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
2
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+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
3
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
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
$${2 \over \lambda } = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$
4
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to :
A
2020 nm
B
250 nm
C
1700 nm
D
220 nm
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