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Chapter 38: Q2P (page 1181)

How fast must an electron move to have a kinetic energy equal to the photon energy of sodium light at wavelength 590 nm?

Short Answer

Expert verified

The speed of the electron is 8.6×105m/s.

Step by step solution

01

Describe the expression for the kinetic energy of the electron and energy of the photon

The expression for kinetic energy of the electron is given by,

K=12mv2

Here, m is the mass of an electron, and v is the speed of the electron.

Combine the above two equations.

E=hcλ

Here, h is the Planck’s constant, c is the speed of the light, and λis the wavelength.

02

Determine the speed of the electron

The energy of the photon is equal to the kinetic energy of the electron.

…… (1)

Substitute below values in equation 1.

m=9.1×10-3Kgλ=590nmh=6.626×10-34J/sc=3×108m/s

Therefore, the speed of the electron is8.6×105m/s

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Most popular questions from this chapter

The wavelength associated with the cutoff frequency for silver is 325nm. Find the maximum kinetic energy of electrons ejected from a silver surface by ultraviolet light of wavelength 254nm.

(a) If you double the kinetic energy of a nonrelativistic particle, how does its de Broglie wavelength change? (b) What if you double the speed of the particle?

Question: Figure 38-13 shows a case in which the momentum component

pxof a particle is fixed so that ∆px=0 ; then, from Heisenberg’s uncertainty principle (Eq. 38-28), 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 (∆x=0), the uncertainty in its momentum is infinite.Consider an intermediate case, in which the position of aparticle is measured, not to infinite precision, but to within a distanceof λ2π, where λis the particle’s de Broglie wavelength.Show that the uncertainty in the (simultaneously measured) momentumcomponent is then equal to the component itself; that is,∆px=p. Under these circumstances, would a measured momentumof zero surprise you? What about a measured momentum of 0.5p? Of 0.2p? Of 12p?

Using the classical equations for momentum and kinetic energy, show that an electron’s de Broglie wavelength in nanometres can be written asλ=1.226/K, in which Kis the electron’s kinetic energy in electron-volts.

In the photoelectric effect (for a given target and a given frequency of the incident light), which of these quantities, if any, depending on the intensity of the incident light beam: (a) the maximum kinetic energy of the electrons, (b) the maximum photoelectric current, (c) the stopping potential, (d) the cut-off frequency?
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