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Q21P

Expert-verifiedFound in: Page 1182

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**X rays with a wavelength of 71 pm are directed onto a gold foil and eject tightly bound electrons from the gold atoms. The ejected electrons then move in circular paths of radius r in a region of the uniform magnetic field $\overrightarrow{{\mathbf{B}}}$. For the fastest of the ejected electrons, the product Br is equal to localid="1664288408568" ${\mathbf{1}}{\mathbf{.}}{\mathbf{88}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{4}}{\mathbf{T}}{\mathbf{\xb7}}{\mathbf{m}}$. Find (a) the maximum kinetic energy of those electrons and (b) the work done in removing them from the gold atoms.**

1. The maximum kinetic energy is 3.1 keV.

2. The work done is 14 keV.

The given data is listed below.

The product of B and r is $\mathrm{B}\times \mathrm{r}=1.88\times {10}^{-4}\mathrm{T}.\mathrm{m}$

The wavelength of the X rays is $\lambda =71pm$

**The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed v. Therefore, kinetic energy on the particle is given by-**

${\mathrm{K}}{=}\frac{1}{2}{\mathrm{m}}{{v}}^{{2}}$

The kinetic energy is a scalar, always positive or zero.

The radius is given by

$r=\frac{{m}_{e}v}{qB}$ ….. (1)

Here, ${m}_{e}$is the mass of the electron, q is the charge of the electron, v is the speed of an electron, and B is the magnetic field.

Now, the speed of an electron can be obtained from equation (1)

$v=\frac{rBq}{{m}_{e}}$

Therefore, kinetic energy is given by-

$\begin{array}{c}{K}_{\mathrm{max}}=\frac{1}{2}{m}_{e}{v}^{2}\\ =\frac{1}{2}{m}_{e}{\left(\frac{rBq}{{m}_{e}}\right)}^{2}\\ =\frac{{\left(rB\right)}^{2}{q}^{2}}{2{m}_{e}}\end{array}$

Thus, the maximum kinetic energy of the electrons is 3.1 keV.

Write the equation for energy of photon as below

${E}_{photon}=\frac{hc}{\lambda}$

Here,

The Plank’s constant is

$\mathrm{h}=6.626\times {10}^{-34}{\mathrm{m}}^{2}.\mathrm{kg}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=\frac{6.626\times {10}^{-34}{\mathrm{m}}^{2}.\mathrm{kg}/\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}}\phantom{\rule{0ex}{0ex}}=4.141\times {10}^{-15}\mathrm{eV}.\mathrm{s}$

The speed of light is,

$\mathrm{c}=3\times {10}^{8}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=3\times {10}^{17}\mathrm{nm}/\mathrm{s}$

Therefore, the value of hc is,

Hence, the work done in removing the electrons from gold atoms is 14 keV.

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