JEE Main 2018 (Online) 16th April Morning Slot | Dual Nature of Radiation Question 141 | Physics

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) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths ^{$$\lambda $$}N, ^{$$\lambda $$}A respectively. The ratio $${{{}^\lambda N} \over {{}^\lambda A}}$$is closest to :

10^$$-$$6

10^$$-$$10

10^$$-$$1

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 masses :

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

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

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

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

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