JEE Main 2019 (Online) 9th January Morning Slot | Dual Nature of Radiation Question 98 | Physics

JEE Main 2019 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)

-1

Surface of certain metal is first illuminated with light of wavelength $$\lambda $$₁ = 350 nm and then, by light of wavelength $$\lambda $$₂ = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :

(Energy of photon n = $${{1240} \over {\lambda (in\,mm)}}$$eV)

1.8

2.5

5.6

1.4

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

The de-Broglie wavelength ($$\lambda $$_B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($$\lambda $$_G) by :

$$\lambda $$_B = 2$$\lambda $$_G

$$\lambda $$_B = 3$$\lambda $$_G

$$\lambda $$_B = $$\lambda $$_G/2

$$\lambda $$_B = $$\lambda $$_G/3

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

-1

If the de Broglie wavelengths associated with a proton and an $$\alpha $$-particle are equal, then the ratio of velocities of the proton and the $$\alpha $$-particle will be :

4 : 1

2 : 1

1 : 2

1 : 4

JEE Main 2018 (Online) 15th April Morning Slot

MCQ (Single Correct Answer)

-1

Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are $${\lambda _1}$$ and $${\lambda _2},$$ their de Broglie wavelength in the frame of reference attached to their center of massis :

$${\lambda _{CM}} = {\lambda _1} = {\lambda _2}$$

$${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$$

$${1 \over {{\lambda _{CM}}}} = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$

$${\lambda _{CM}} = \left( {{{{\lambda _1} + {\lambda _2}} \over 2}} \right)$$

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