1
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
If the de Broglie wavelength of an electron is equal to the 10–3 times the wavelength of a photon of frequency 6 $$\times$$ 1014 Hz, then the speed of electron is equal to : (Speed of light = 3 $$\times$$ 108 m/s, Planck's constant = 6.63 $$\times$$ 10–34 J.s, Mass of electron = 9.1 $$\times$$ 10–31 kg)
A
1.7 $$\times$$ 106 m/s
B
1.45 $$\times$$ 106 m/s
C
1.1 $$\times$$ 106 m/s
D
1.8 $$\times$$ 106 m/s
2
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
A metal plate of area 1 $$\times$$ 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photoelectrons per second and their maximum energy, respectively, will be
A
1014 and 10 eV
B
1012 and 5 eV
C
1011 and 5 eV
D
1010 and 5 eV
3
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
4
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
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