The ratio of the power of a light source S1 to that of the light source S2 is 2. S1 is emitting 2 1015 photons per second at 600 nm. If the wavelength of the source S2 is 300 nm, then the number of photons per second emitted by S2 is 1014.
5
Power P=n hc n P
n2=n1 P2P1 2 1
=2 1015 12 300600
=5 1014
Answer: 5
A sub-atomic particle of mass 10-30 kg is moving with a velocity 2.21 106 m/s. Under the matter wave consideration, the particle will behave closely like (h=6.63 10-34 J.s )
In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V, what is the wavelength of the em-wave? (Given hc=1242 eV nm where h is the Planck's constant and c is the speed of light in vacuum.)
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