Chapter 18, Problem 3SP

(a)

To determine

The energy required to ionize a hydrogen atom in its lowest energy level.

Answer to Problem 3SP

The energy required to ionize a hydrogen atom in its lowest energy level is $13.6\text{ eV}$.

Explanation of Solution

The energy level diagram of hydrogen proposed by the Bohr model gives the energy of electrons in different orbits in hydrogen atom. The energy of the ground state or the lowest level in the hydrogen atom is $-13.6\text{ eV}$.

Ionization energy is the energy required to remove an electron from its orbit. It will be equal to the negative of the energy of the electron in the orbit. The energy of the electron in the lowest level of hydrogen atom is $-13.6\text{ eV}$. The negative of $-13.6\text{ eV}$ is $13.6\text{ eV}$.

Conclusion:

Thus the energy required to ionize a hydrogen atom in its lowest energy level is $13.6\text{ eV}$.

(b)

To determine

The energy required to ionize the hydrogen atom when it is in the first excited state above the lowest level.

Answer to Problem 3SP

The energy required to ionize the hydrogen atom when it is in the first excited state above the lowest level is $3.4\text{ eV}$.

Explanation of Solution

The energy level diagram of hydrogen proposed by the Bohr model gives the energy of electrons in different orbits in hydrogen atom. The energy of a particular level in the hydrogen atom is given by the ratio of $-13.6\text{ eV}$ to the square of the principal quantum number of the particular energy level.

The first excited state is the level which lies just above the ground level. The energy of the first excited state of hydrogen atom is $-3.4\text{ eV}$.

Ionization energy is the energy required to remove an electron from its orbit. It will be equal to the negative of the energy of the electron in the orbit. This implies the energy required to ionize the hydrogen atom in the first excited state will be equal to $3.4\text{ eV}$.

Conclusion:

Thus the energy required to ionize the hydrogen atom when it is in the first excited state above the lowest level is $3.4\text{ eV}$.

(c)

To determine

The wavelength of the photon emitted when an ionized hydrogen atom goes to the lowest energy level after capturing an electron with zero kinetic energy.

Answer to Problem 3SP

The wavelength of the photon emitted when an ionized hydrogen atom goes to the lowest energy level after capturing an electron with zero kinetic energy is $91.3\text{ nm}$.

Explanation of Solution

An ionized hydrogen atom will have zero energy. The energy of the lowest level in the hydrogen atom is $-13.6\text{ eV}$. The energy of the photon emitted will be equal to the energy difference between the final and initial states of the atom.

Write the equation for the energy of the photon.

$E=E_f-E_i$

Here,

$E$ is the energy of the photon

$E_f$ is the final energy of the atom

$E_i$ is the initial energy of the atom

Substitute $0\text{ eV}$ for $E_f$ and $-13.6\text{ eV}$ for $E_i$ in the above equation to find $E$.

$\begin{array}{c}E=0\text{ eV}-\left(-13.6\text{ eV}\right)\\ =13.6\text{ eV}\end{array}$

Write the equation for the energy of a photon.

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

Here,

$h$ is the Planck’s constant

$c$ is the speed of light in vacuum

Rewrite the above equation for $\lambda$.

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

The value of $h$ is $6.626\times {10}^{-34}\text{ J}\cdot \text{s}$ and the value of $c$ is $3.0\times {10}^8\text{ m/s}$.

Substitute $6.626\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$,$3.0\times {10}^8\text{ m/s}$ for $c$ and $13.6\text{ eV}$ for $E$ in the above equation to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{ m/s}\right)}{13.6\text{ eV}\left(\frac{1.60\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)}\\ =\frac{\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{ m/s}\right)}{2.176\times {10}^{-18}\text{ J}}\\ =91.3\times {10}^{-9}\text{ m}\\ =91.3\text{ nm}\end{array}$

Conclusion:

Thus the wavelength of the photon emitted when an ionized hydrogen atom goes to the lowest energy level after capturing an electron with zero kinetic energy is $91.3\text{ nm}$