1
JEE Main 2018 (Online) 16th April Morning Slot
+4
-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 :
A
$$\lambda$$B = 2$$\lambda$$G
B
$$\lambda$$B = 3$$\lambda$$G
C
$$\lambda$$B = $$\lambda$$G/2
D
$$\lambda$$B = $$\lambda$$G/3
2
JEE Main 2018 (Online) 16th April Morning Slot
+4
-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 :
A
10$$-$$6
B
10
C
10$$-$$10
D
10$$-$$1
3
JEE Main 2018 (Online) 15th April Evening Slot
+4
-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 :
A
4 : 1
B
2 : 1
C
1 : 2
D
1 : 4
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-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 :
A
$${\lambda _{CM}} = {\lambda _1} = {\lambda _2}$$
B
$${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$$
C
$${1 \over {{\lambda _{CM}}}} = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}$$
D
$${\lambda _{CM}} = \left( {{{{\lambda _1} + {\lambda _2}} \over 2}} \right)$$
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