1
JEE Main 2017 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $${\lambda _A}$$ to $${\lambda _B}$$ after the collision is:
A
$${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 3}$$
B
$${{{\lambda _A}} \over {{\lambda _B}}} = 2$$
C
$${{{\lambda _A}} \over {{\lambda _B}}} = {2 \over 3}$$
D
$${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 2}$$
2
JEE Main 2016 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A photoelectric surface is illuminated successively by monochromatic light of wavelengths $$\lambda $$ and $${\lambda \over 2}.$$ If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :
A
$${{hc} \over {3\lambda }}$$
B
$${{hc} \over {2\lambda }}$$
C
$${{hc} \over {\lambda }}$$
D
$${3\,{hc} \over {\lambda }}$$
3
JEE Main 2016 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
When photons of wavelength $${\lambda _1}$$ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength $${\lambda _2}$$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength $${\lambda _3}$$ is used then find the stopping potential for this case :
A
$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} - {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$
B
$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$
C
$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {3 \over {2{\lambda _1}}}} \right]$$
D
$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$
4
JEE Main 2016 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
Radiation of wavelength $$\lambda ,$$ is incident on a photocell. The fastest emitted electron has speed $$v.$$ If the wavelength is changed to $${{3\lambda } \over 4},$$ the speed of the fastest emitted electron will be:
A
$$ = v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$
B
$$ = v{\left( {{3 \over 4}} \right)^{{1 \over 2}}}$$
C
$$ > v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$
D
$$ < v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$$
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