JEE Main 2020 (Online) 9th January Evening Slot | Dual Nature of Radiation Question 118 | Physics

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 :

$${{ - h} \over {\left| e \right|Et}}$$

$${{ - h} \over {\left| e \right|E\sqrt t }}$$

$${{ - h} \over {\left| e \right|E{t^2}}}$$

$${{\left| e \right|Et} \over h}$$

JEE Main 2020 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)

-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 :

JEE Main 2020 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)

-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 :

1.8eV

0.8eV

1.1eV

1.6eV

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 :

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

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

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

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

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