1
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
2
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 × 10–4 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
3
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
$${{{\lambda _0}} \over {\sqrt {2 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$
4
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength $$\lambda _A$$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons $$\lambda _B$$ = 2$$\lambda _A$$, then the work function of metal B is :
A
1.5eV
B
4eV
C
2eV
D
3eV
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