The momentum of a photon of energy $h\nu $ will be
The momentum of a photon of energy $h\nu $ will be
- A
$h\nu $
- B
$\frac{{h\nu}}{{c}} $
- C
$h\nu c$
- D
$h/\nu $
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The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600\, \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3}$ $watt$ will be : $\left(\mathrm{h}=6.6 \times 10^{-34}\, \mathrm{Js}\right)$
The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600\, \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3}$ $watt$ will be : $\left(\mathrm{h}=6.6 \times 10^{-34}\, \mathrm{Js}\right)$
- [NEET 2021]
When radiation of wavelength $\lambda $ is incident on a metallic surface, the stopping potential is $4.8\, volts$. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes $1.6\, volts$. Then the threshold wavelength for the surface is
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A radiation of energy $'E'$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $( C =$ Velocity of light $)$
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- [AIEEE 2004]
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The photoelectric effect can be understood on the basis of
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