JEE Main 2021 (Online) 20th July Evening Shift | Dual Nature of Radiation Question 48 | Physics

JEE Main 2021 (Online) 20th July Evening Shift

MCQ (Single Correct Answer)

-1

An electron having de-Broglie wavelength $$\lambda$$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is :

$${{2{m^2}{c^2}{\lambda ^2}} \over {{h^2}}}$$

$${{2mc{\lambda ^2}} \over h}$$

$${{hc} \over {mc}}$$

JEE Main 2021 (Online) 20th July Morning Shift

MCQ (Single Correct Answer)

-1

The radiation corresponding to 3 $$\to$$ 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 $$\times$$ 10^$$-$$4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron = 9.1 $$\times$$ 10^$$-$$31 kg)

1.36 eV

1.88 eV

0.82 eV

0.16 eV

JEE Main 2021 (Online) 17th March Evening Shift

MCQ (Single Correct Answer)

-1

Two identical photocathodes receive the light of frequencies f₁ and f₂ respectively. If the velocities of the photo-electrons coming out are v₁ and v₂ respectively, then

$${v_1} - {v_2} = {\left[ {{{2h} \over m}({f_1} - {f_2})} \right]^{{1 \over 2}}}$$

$$v_1^2 + v_2^2 = {{2h} \over m}[{f_1} + {f_2}]$$

$${v_1} + {v_2} = {\left[ {{{2h} \over m}({f_1} + {f_2})} \right]^{{1 \over 2}}}$$

$$v_1^2 - v_2^2 = {{2h} \over m}[{f_1} - {f_2}]$$

JEE Main 2021 (Online) 16th March Morning Shift

MCQ (Single Correct Answer)

-1

The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation :

Phase

Frequency

Amnplitude

Intensity