Consider figure for photoemission.
How would you reconcile with momentum-conservation ? Note light (photons) have momentum in a different direction than the emitted electrons.
Consider figure for photoemission.
How would you reconcile with momentum-conservation ? Note light (photons) have momentum in a different direction than the emitted electrons.
When incident photon is absorbed completely, its momentum becomes zero. This momentum is gained by atoms of metal and so these atoms get excited. As a result, electrons in them, transit from lower to higher energy levels and along with this, some free electrons get emitted in the form of photoelectrons. Here, total momentum is definitely conserved, irrespective of type of collision.
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A $2\,mW$ laser operates at a wavelength of $500\,nm.$ The number of photons that will be emitted per second is [Given Planck’s constant $h = 6.6 \times 10^{-34}\,Js,$ speed of light $c = 3.0\times 10^8\,m/s$ ]
A $2\,mW$ laser operates at a wavelength of $500\,nm.$ The number of photons that will be emitted per second is [Given Planck’s constant $h = 6.6 \times 10^{-34}\,Js,$ speed of light $c = 3.0\times 10^8\,m/s$ ]
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Two monochromatic beams $A$ and $B$ of equal intensity $I$, hit a screen. The number of photons hitting the screen by beam $A $ is twice that by beam $ B$. Then what inference can you make about their frequencies ?
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In a dark room of photography, generally red light is used. The reason is
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