JEE Main 2020 (Online) 2nd September Morning Slot | Atoms and Nuclei Question 117 | Physics

JEE Main 2020 (Online) 2nd September Morning Slot

MCQ (Single Correct Answer)

-1

In a reactor, 2 kg of ₉₂U²³⁵ fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N = 6.023 $$ \times $$ 10²⁶ per kilo mole and 1 eV = 1.6 × 10^–19 J. The power output of the reactor is close to

125 MW

60 MW

54 MW

35 MW

JEE Main 2020 (Online) 8th January Evening Slot

MCQ (Single Correct Answer)

-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 :

$${{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$

$${{{\lambda _0}\sqrt 2 } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$

$${{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {2{m^2}v_0^2}}} }}$$

$${{{\lambda _0}} \over {\sqrt {2 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$

JEE Main 2020 (Online) 8th January Morning Slot

MCQ (Single Correct Answer)

-1

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_A 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 T_B = (T_A – 1.5) eV. If the de-Broglie wavelength of these photoelectrons $$\lambda _B$$ = 2$$\lambda _A$$, then the work function of metal B is :

1.5eV

4eV

2eV

3eV

JEE Main 2020 (Online) 8th January Morning Slot

MCQ (Single Correct Answer)

-1

The graph which depicts the results of Rutherford gold foil experiment with $$\alpha $$-particales is :
$$\theta $$ : Scattering angle
Y : Number of scattered $$\alpha $$-particles detected (Plots are schematic and not to scale)