JEE Main 2021 (Online) 20th July Morning Shift | Dual Nature of Radiation Question 112 | Physics

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) 18th March Evening Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

The speed of electrons in a scanning electron microscope is 1 $$\times$$ 10⁷ ms^-1. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of :

$${1 \over {1837}}$$

1837

$${\sqrt {1837} }$$

$${1 \over {\sqrt {1837} }}$$

JEE Main 2021 (Online) 18th March Morning Shift

MCQ (Single Correct Answer)

-1

A particle is travelling 4 time as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2 : 1, the mass of the particle is :

$${1 \over {16}}$$ times the mass of e^$$-$$

8 times the mass of e^$$-$$

16 times the mass of e^$$-$$

$${1 \over {8}}$$ times the mass of e^$$-$$

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

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

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

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

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

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