1
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
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 ?
(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
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination