Question

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).

Accepted Answer

Given, $h$ is the Planck's constant,

$c$ is the speed of the light,

$λ$ is the wavelength of the light used.

$(a)$ Using the value $h c = 1240 nm eV$ , we have,

$E = \frac{h c}{λ} = \frac{1240 nm eV}{10.0 \times 10^{- 3} nm} = 124 keV$ .

$(b)$ The kinetic energy gained by the electron is equal to the energy decrease of the photon.

$Δ λ$ is the Compton shift which is given by, $Δ λ = \frac{h}{m c} (1 - \cos (Φ))$ , where $m$ is mass of the electron.

$ϕ = 180 °$

$Δ E = h c (\frac{1}{λ} - \frac{1}{λ + Δλ}) = (\frac{h c}{λ}) (\frac{Δλ}{λ + Δλ}) = \frac{E}{1 + \frac{λ}{Δλ}} \begin{array}{l} = Δ E = \frac{124 keV}{\frac{(10 \times 10^{- 12} m) (9 .11 \times 10^{- 31} kg) (2 .998 \times 10^{8} m s^{- 1})}{2 (6 .626 \times 10^{- 34} J s)} + 1} \\ = 40.5 keV \end{array}$

$(c)$ Here, it is impossible to view an atomic electron with such a high-energy photon because with the energy imparted to the electron the photon would have knocked the electron out of its orbit.

Important Questions on Photons and Matter Waves

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