Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Show that the cutoff wavelength in the continuous x-ray spectrum from any target is given by $λ_{\min} = 1240 / V$ , where is the potential difference (in kilovolts) through which the electrons are accelerated before they strike the target.

It is shown that the cutoff wavelength in the continuous x-ray spectrum from any target is given by $λ_{m i n} = 1240 / V$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

A continuous x-ray spectrum from any target is produced due to the striking of the electrons that are accelerating before the strike.

Step 2: Understanding the concept of Plank’s relation:

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Using the energy relation of Planck's equation and the energy difference created by the accelerating electron due to generated potential difference, to get the required equation of the cutoff wavelength in the continuous x-ray spectrum from any target.

Formulae:

The energy of the photon due to Planck’s relation,

$Δ E = \frac{h c}{λ}$ ….. (1)

Consider the known data below.

The Plank’s constant is,

$h = 6.63 \times 10^{- 34} J . s = (6.242 \times 10^{15} \times 6.63 \times 10^{- 34}) keV . s = 41.384 \times 10^{19} keV . s$

The speed of light is,

$c = 3 \times 10^{8} m / s = (3 \times 10^{8} \times 10^{12}) pm / s = 3 \times 10^{20} pm / s$

The energy generated due to accelerating potential,

$Δ E = e V$ ….. (2)

Here, e is the charge and V is the potential.

Step 3: Calculation of the cut-off wavelength:

As the accelerating electrons strike the target, they generate the same energy difference on the target as due to the accelerating potential. Thus, using equations (1) and (2), the cutoff wavelength in the continuous x-ray spectrum from any target is given by:

$e V = \frac{h c}{λ_{m i n}}$

$λ_{m i n} = \frac{h c}{e V} = \frac{(41.384 \times 10^{- 19} k e V . s) (3 \times 10^{20} p m / s)}{e V} = \frac{1240 k e V . p m}{e V} = \frac{1240 p m}{V}$

Here, the potential V is in the kilovolts.

Hence, it is proved that the wavelength value is $λ_{m i n} = \frac{1240}{V}$ .

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Fundamentals Of Physics

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Short Answer

Show that the cutoff wavelength in the continuous x-ray spectrum from any target is given by $λ_{\min} = 1240 / V$ , where is the potential difference (in kilovolts) through which the electrons are accelerated before they strike the target.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of Plank’s relation:

Step 3: Calculation of the cut-off wavelength:

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Fundamentals Of Physics

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Short Answer

Show that the cutoff wavelength in the continuous x-ray spectrum from any target is given by λmin=1240/V, where is the potential difference (in kilovolts) through which the electrons are accelerated before they strike the target.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of Plank’s relation:

Step 3: Calculation of the cut-off wavelength:

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Show that the cutoff wavelength in the continuous x-ray spectrum from any target is given by $λ_{\min} = 1240 / V$ , where is the potential difference (in kilovolts) through which the electrons are accelerated before they strike the target.