Chapter 28, Problem 45P

To determine

The ionization energies of L, M and N shell.

Answer to Problem 45P

The ionization energies of L, M and N shell is $11.7\text{keV , 10 keV , 2,3 keV}$ respectively.

Explanation of Solution

Section1:

To determine: The ionization energy of the L-shell.

Answer: The ionization energy of the L-shell is $11.7\text{keV}$.

Explanation:

Given Info: The ionization energy of K-shell is $69.5\text{keV}$.

Formula to calculate the ionization energy of the L-shell is,

$E_L=E_k+\frac{hc}{{\lambda }_3}$

• $E_L$ is ionization energy of the L-shell,
• $E_K$ is ionization energy of the K-shell,
• $h$ is the Planck’s Constant
• $c$ is the speed of light
• ${\lambda }_3$ is the wavelength

From unit conversion,

$1\text{keV}=1.6\times {10}^{-16}\text{J}$

Substitute $6.63\times {10}^{-34}\text{J-s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $0.0215\text{nm}$ for ${\lambda }_3$, $\left(-69.5keV\right)$ for $E_K$ to find $E_L$,

$\begin{array}{c}{E}_L=\left(-69.5\text{keV}\right)+\left[\frac{\left(6.63\times {10}^{-34}\text{J-s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(0.0215\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)}\left(\frac{1\text{keV}}{1.6\times {10}^{-16}\text{J}}\right)\right]\\ =-11.7\text{keV}\end{array}$

Thus, ionization energy of the L-shell electron is $11.7\text{keV}$.

Section 2:

To determine: The ionization energy of the M-shell.

Answer: The ionization energy of the M-shell is $10\text{keV}$.

Explanation:

Given Info: The ionization energy of K-shell is $69.5\text{keV}$.

Formula to calculate the ionization energy of the M-shell is,

$E_M=E_k+\frac{hc}{{\lambda }_2}$

• $E_M$ is ionization energy of the M-shell,
• $E_K$ is ionization energy of the K-shell,
• $h$ is the Planck’s Constant
• $c$ is the speed of light
• ${\lambda }_2$ is the wavelength

From unit conversion,

$1\text{keV}=1.6\times {10}^{-16}\text{J}$

Substitute $6.63\times {10}^{-34}\text{J-s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $0.0209\text{nm}$ for ${\lambda }_2$, $\left(-69.5keV\right)$ for $E_K$ to find $E_M$,

$\begin{array}{c}{E}_M=\left(-69.5\text{keV}\right)+\left[\frac{\left(6.63\times {10}^{-34}\text{J-s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(0.0209\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)}\left(\frac{1\text{keV}}{1.6\times {10}^{-16}\text{J}}\right)\right]\\ =-10\text{keV}\end{array}$

Thus, ionization energy of the M-shell electron is $10\text{keV}$.

Section3:

To determine: The ionization energy of the N-shell.

Answer: The ionization energy of the N-shell is $2.3\text{keV}$.

Explanation:

Given Info: The ionization energy of K-shell is $69.5\text{keV}$.

Formula to calculate the ionization energy of the N-shell is,

$E_N=E_k+\frac{hc}{{\lambda }_1}$

• $E_N$ is ionization energy of the L-shell,
• $E_K$ is ionization energy of the K-shell,
• $h$ is the Planck’s Constant
• $c$ is the speed of light
• ${\lambda }_1$ is the wavelength

From unit conversion,

$1\text{keV}=1.6\times {10}^{-16}\text{J}$

Substitute $6.63\times {10}^{-34}\text{J-s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $0.0185\text{nm}$ for ${\lambda }_1$, $\left(-69.5keV\right)$ for $E_K$ to find $E_N$,

$\begin{array}{c}{E}_N=\left(-69.5\text{keV}\right)+\left[\frac{\left(6.63\times {10}^{-34}\text{J-s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(0.0185\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)}\left(\frac{1\text{keV}}{1.6\times {10}^{-16}\text{J}}\right)\right]\\ =-2.3\text{keV}\end{array}$

Thus, ionization energy of the N-shell electron is $2.3\text{keV}$.

Conclusion:

Therefore, the ionization energies of L, M and N shell is $11.7\text{keV , 10 keV , 2,3 keV}$ respectively.