Chapter 29, Problem 38P

(a)

To determine

The ionization energies of the L, M and N shells.

Answer to Problem 38P

The ionization energies for the L shell is $11.8\text{keV}$, M shell is $10.2\text{keV}$ and N shells is $2.47\text{keV}$

Explanation of Solution

The figure 1 shows the transitions from higher levels N shell down to K shell.

The K series includes the transition from higher levels down to the K shell $\left(n=1\right)$. Transition from higher state produces photons of higher energy.

The ionization energy for the K shell is $69.5\text{keV}$, so the energy of the K shell is $-69.5\text{keV}$

Write the expression for energy of the photon.

  $E=\frac{hc}{\lambda }$        (I)

Here, $E$ is the energy of the photon, $h$ is the plank’s constant, $c$ is the velocity of light, and $\lambda$ is the wavelength.

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, and $0.0185\text{nm}$ for ${\lambda }_1$ in the equation (I) to find $E_1$.

  $\begin{array}{c}{E}_1=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(0.0185\text{nm}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(0.0185\text{nm}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{0.0185\text{nm}}\\ =67.03\times {10}^3\text{eV}\end{array}$

Convert the energy of the photon into kilo electron volt.

  $\begin{array}{c}{E}_1=67.03\times {10}^3\text{eV}\left(\frac{{10}^{-3}\text{keV}}{1\text{eV}}\right)\\ =67.03\text{keV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, and $0.0209\text{nm}$ for ${\lambda }_2$ in the equation (I) to find $E_2$.

  $\begin{array}{c}{E}_2=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(0.0209\text{nm}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(0.0209\text{nm}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{0.0209\text{nm}}\\ =59.3\times {10}^3\text{eV}\end{array}$

Convert the energy of the photon into kilo electron volt.

  $\begin{array}{c}{E}_2=59.3\times {10}^3\text{eV}\left(\frac{{10}^{-3}\text{keV}}{1\text{eV}}\right)\\ =59.3\text{keV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, and $0.0215\text{nm}$ for ${\lambda }_3$ in the equation (I) to find $E_3$.

  $\begin{array}{c}{E}_3=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(0.0215\text{nm}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(0.0215\text{nm}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{0.0215\text{nm}}\\ =57.7\times {10}^3\text{eV}\end{array}$

Convert the energy of the photon into kilo electron volt.

  $\begin{array}{c}{E}_3=57.7\times {10}^3\text{eV}\left(\frac{{10}^{-3}\text{keV}}{1\text{eV}}\right)\\ =57.7\text{keV}\end{array}$

Similarly the table 1 shows the ionization energy for the shells.

$\lambda \left(\text{nm}\right)$	Photon energy $\left(\text{keV}\right)$	transition	Energy of level	level
${\lambda }_1=0.0185$	67.03	$n=4\to 1$	$-69.5\text{keV}+67.03\text{keV}=-2.47\text{keV}$	N
${\lambda }_2=0.0209$	59.3	$n=3\to 1$	$-69.5\text{keV}+59.3\text{keV}=-10.2\text{keV}$	M
${\lambda }_3=0.0215$	57.7	$n=2\to 1$	$-69.5\text{keV}+57.7\text{keV}=-11.8\text{keV}$	L

The ionization energy for the K shell is $69.5\text{keV}$, so the ionization energies for other shell.

Therefore, the ionization energies for the L shell is $11.8\text{keV}$, M shell is $10.2\text{keV}$ and N shells is $2.47\text{keV}$

(b)

To determine

Draw the diagram of the transition.

Answer to Problem 38P

The table 1 shows the transition of the X-ray spectrum of tungsten element.

Explanation of Solution

From (a) The table shows the transition of the X ray spectrum..

$\lambda \left(\text{nm}\right)$	Photon energy $\left(\text{keV}\right)$	transition	Energy of level	level
${\lambda }_1=0.0185$	67.03	$n=4\to 1$	$-69.5\text{keV}+67.03\text{keV}=-2.47\text{keV}$	N
${\lambda }_2=0.0209$	59.3	$n=3\to 1$	$-69.5\text{keV}+59.3\text{keV}=-10.2\text{keV}$	M
${\lambda }_3=0.0215$	57.7	$n=2\to 1$	$-69.5\text{keV}+57.7\text{keV}=-11.8\text{keV}$	L

Conclusion:

Therefore, the table 1 shows the transition of the X-ray spectrum of tungsten element..