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Figure shows a case in which the momentum component px of a particle is fixed, so that, Δpx=0; then, from Heisenberg's uncertainty principle, the position x of the particle is completely unknown. From the same principle, it follows that the opposite is also true, that is, if the position of a particle is exactly known (Δx=0), the uncertainty in its momentum is infinite.

Consider an intermediate case. In which the position of a particle is measured, not to infinite precision, but to within a distance of λ/2π, where λ is the particle's de Broglie wavelength. Show that the uncertainty in the (simultaneously measured) momentum component is then equal to the component itself; that is, Δpx=p. Under these circumstances, would a measured momentum of zero surprises you? What about a measured momentum of 0.5p? Of 2p? Of 12p?

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Important Questions on Photons and Matter Waves

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IMPORTANT
The uncertainty in the position of an electron along the x-axis is given as 50 pm, which is about equal to the radius of a hydrogen atom. What is the least uncertainty in any simultaneous measurement of the momentum component px of this electron?
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IMPORTANT
Suppose the fractional efficiency of a cesium surface(with work function 1.80 eV) is 1.0×10-16; that is, on average one electron is ejected for every 1016 photon that reaches the surface. What would be the current of electrons ejected from such a surface if it were illuminated with 600 nm light from a 3.00 mW laser and all the ejected electrons took part in the charge flow?
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JEE Main
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An electron and a photon each have a wavelength of 0.25 nm. What is the momentum in kg m s-1 of the (a) electron and (b) photon? What is the energy in eV of the (c) electron and (d) photon?
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IMPORTANT

The function ψ(x) displayed in Eq. 38-27 can describe a free particle for which the potential energy is U(x)=0 in Schrodinger's equation (Eq. 38-19). Assume now that U(x)=U0= a constant in that equation. Show that Eq. 38-27 is a solution of Schrodinger's equation with

k=2πh2mE-U0

giving the angular wave number k of the particle.

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JEE Main
IMPORTANT
Show that the angular wave number k for a non-relativistic free particle of mass m is k=2π2mKh, can be written as, in which K is the particle's kinetic energy.
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JEE Main
IMPORTANT
You will find that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to observe such an orbiting electron by using a light microscope to measure the electron's presumed orbital position with a precision of, say, 10 pm (a typical atom has a radius of about 100 pm). The wavelength of the light used in the microscope must then be about 10 pm(a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of viewing an atomic electron at two or more points along its presumed orbital path? (Hint: The outer electrons of atoms are bound to the atom by energies of only a few electron-volts). 
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An ultraviolet lamp emits light of wavelength 400 nm at the rate of 400 W. An infrared lamp emits light of wavelength 700 nm, also at the rate of 400 W. (a) Which lamp emits photons at the greater rate and (b) what is that greater rate?
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The sun is approximately an ideal blackbody radiator with a surface temperature of 5800 K. (a) Find the wavelength at which its spectral radiancy is maximum and (b) identify the type of electromagnetic wave corresponding to that wavelength. (See Figure) (c) The universe is approximately an ideal blackbody radiator with radiation emitted when atoms first formed. Today the spectral radiancy of that radiation peaks at a wavelength of 1.06 mm (in the microwave region). What is the corresponding temperature of the universe?

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