1
JEE Main 2020 (Online) 9th January Evening Slot
+4
-1
An electron of mass m and magnitude of charge |e| initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is :
A
$${{ - h} \over {\left| e \right|Et}}$$
B
$${{ - h} \over {\left| e \right|E\sqrt t }}$$
C
$${{ - h} \over {\left| e \right|E{t^2}}}$$
D
$${{\left| e \right|Et} \over h}$$
2
JEE Main 2020 (Online) 9th January Morning Slot
+4
-1
A particle moving with kinetic energy E has de Broglie wavelength $$\lambda$$. If energy $$\Delta$$E is added to its energy, the wavelength become $$\lambda$$/2. Value of $$\Delta$$E, is :
A
E
B
3E
C
2E
D
4E
3
JEE Main 2020 (Online) 9th January Morning Slot
+4
-1
Radiation, with wavelength 6561 $$\mathop A\limits^o$$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 Ć 10ā4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
A
1.8eV
B
0.8eV
C
1.1eV
D
1.6eV
4
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
An electron (mass m) with initial velocity $$\overrightarrow v = {v_0}\widehat i + {v_0}\widehat j$$ is in an electric field $$\overrightarrow E = - {E_0}\widehat k$$. If $$\lambda _0$$ is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :
A
$${{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$
B
$${{{\lambda _0}\sqrt 2 } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$
C
$${{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {2{m^2}v_0^2}}} }}$$
D
$${{{\lambda _0}} \over {\sqrt {2 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$
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