1
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -
A
25 keV
B
500 keV
C
100 keV
D
1 keV
2
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 $$\times$$ 107)ct + sin(6.28 $$\times$$ 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(Take c = 3 $$\times$$ 108 ms$$-$$1, h = 6.6 $$\times$$ 10$$-$$34J-s)
A
6.82 eV
B
12.5 eV
C
8.52 eV
D
7.72 eV
3
JEE Main 2019 (Online) 9th January Morning Slot
+4
-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)
A
1.8
B
2.5
C
5.6
D
1.4
4
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
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