Chapter 29, Problem 45P

(a)

To determine

The energies of ground state, and first four excited states.

Answer to Problem 45P

The energies of the ground state is $-8.16\text{eV}$ and first four excited states.is $E_2=-2.04\text{eV, }{E}_3=-0.902\text{eV, }{E}_4=-0.508\text{eV, and }{E}_5=-0.325\text{eV}$.

Explanation of Solution

Write the expression for the energy of the ground state.

  $E_1=-\frac{hc}{{\lambda }_{\text{series limit}}}$        (I)

Here, $E_1$ is the photon’s energy from its ground state, $c$ is the velocity of light, ${\lambda }_{\text{series limit}}$ is the wavelength of the series of spectral line, and $h$ is the plank’s constant.

Write the expression for the energy of the excited state of Lyman $\alpha$ line.

  $E_2-E_1=\frac{hc}{{\lambda }_1}$        (II)

Here, $E_2$ is the energy of the electron from the excited state of Lyman $\alpha$ line and ${\lambda }_{\text{1}}$ is the wavelength of the of Lyman $\alpha$ line.

Write the expression for the energy of the excited state of Lyman $\beta$ line.

  $E_3-E_1=\frac{hc}{{\lambda }_2}$        (III)

Here, $E_3$ is the energy of the electron from the excited state of Lyman $\beta$ line and ${\lambda }_2$ is the wavelength of the of Lyman $\beta$ line.

Write the expression for the energy of the excited state of Lyman $\gamma$ line.

  $E_4-E_1=\frac{hc}{{\lambda }_3}$        (IV)

Here, $E_4$ is the energy of the electron from the excited state of Lyman $\gamma$ line and ${\lambda }_3$ is the wavelength of the of Lyman $\gamma$ line.

Write the expression for the energy of the excited state of Lyman $\delta$ line.

  $E_5-E_1=\frac{hc}{{\lambda }_4}$        (V)

Here, $E_5$ is the energy of the electron from the excited state of Lyman $\delta$ line and ${\lambda }_4$ is the wavelength of the of Lyman $\delta$ line.

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, and $152.0\text{nm}$ for ${\lambda }_{\text{series limit}}$ in the equation (I) to find $E_1$.

  $\begin{array}{c}{E}_1=-\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(152.0\text{nm}\right)}\\ =-\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(152.0\text{nm}\right)}\\ =-\frac{1240\text{eV}\cdot \text{nm}}{152.0\text{nm}}\\ =-8.16\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $-8.16\text{eV}$ for $E_1$, and $202.6\text{nm}$ for ${\lambda }_1$ in the equation (II) to find $E_2$.

  $\begin{array}{c}{E}_2=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(202.6\text{nm}\right)}+E_1\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(202.6\text{nm}\right)}-8.16\text{eV}\\ =\frac{1240\text{eV}\cdot \text{nm}}{202.6\text{nm}}-8.16\text{eV}\\ =-2.04\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $-8.16\text{eV}$ for $E_1$, and $170.9\text{nm}$ for ${\lambda }_2$ in the equation (III) to find $E_3$.

  $\begin{array}{c}{E}_3=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(170.9\text{nm}\right)}+E_1\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(170.9\text{nm}\right)}-8.16\text{eV}\\ =\frac{1240\text{eV}\cdot \text{nm}}{170.9\text{nm}}-8.16\text{eV}\\ =-0.902\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $-8.16\text{eV}$ for $E_1$, and $162.1\text{nm}$ for ${\lambda }_3$ in the equation (IV) to find $E_4$.

  $\begin{array}{c}{E}_4=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(162.1\text{nm}\right)}+E_1\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(162.1\text{nm}\right)}-8.16\text{eV}\\ =\frac{1240\text{eV}\cdot \text{nm}}{162.1\text{nm}}-8.16\text{eV}\\ =-0.508\text{eV}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $-8.16\text{eV}$ for $E_1$, and $158.3\text{nm}$ for ${\lambda }_4$ in the equation (V) to find $E_5$.

  $\begin{array}{c}{E}_5=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(158.3\text{nm}\right)}+E_1\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(158.3\text{nm}\right)}-8.16\text{eV}\\ =\frac{1240\text{eV}\cdot \text{nm}}{158.3\text{nm}}-8.16\text{eV}\\ =-0.325\text{eV}\end{array}$

Therefore, The energies of the ground state is $-8.16\text{eV}$ and first four excited states.is $E_2=-2.04\text{eV, }{E}_3=-0.902\text{eV, }{E}_4=-0.508\text{eV, and }{E}_5=-0.325\text{eV}$.

(b)

To determine

The wavelength of the first three lines and one short wavelength limit in Balmer series.

Answer to Problem 45P

The wavelength of the first three lines is ${\lambda }_{\alpha }=1090\text{nm}$, ${\lambda }_{\beta }=811\text{nm}$, ${\lambda }_{\alpha }=724\text{nm}$ and one short wavelength limit in Balmer series is $609\text{nm}$.

Explanation of Solution

Write the expression for energy of the Balmer series.

  $E_i-E_2=\frac{hc}{\lambda }$

Here, $E_i$ is the ionization energy of the series.

For the $\alpha$ spectral line, substitute $E_3$ for $E_i$ and ${\lambda }_{\alpha }$ for $\lambda$ in the above expression.

  $E_3-E_2=\frac{hc}{{\lambda }_a}$

Rewrite the above expression for ${\lambda }_{\alpha }$.

  ${\lambda }_{\alpha }=\frac{hc}{E_3-E_2}$        (VI)

Similarly the expression for the wavelength of the $\beta$ line is,

  ${\lambda }_{\beta }=\frac{hc}{E_4-E_2}$        (VII)

Similarly the expression for the wavelength of the $\gamma$ line is,

  ${\lambda }_{\gamma }=\frac{hc}{E_5-E_2}$        (VIII)

Similarly the expression for the short wavelength limit is,

  ${\lambda }_{\text{short}}=\frac{hc}{E_1-E_2}$        (IX)

Conclusion:

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $-2.04\text{eV}$ for $E_2$, and $-0.902\text{eV}$ for $E_3$ in the equation (VI) to find ${\lambda }_{\alpha }$

  $\begin{array}{c}{\lambda }_{\alpha }=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(-0.902\text{eV}\right)-\left(-2.04\text{eV}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(1.138\text{eV}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{1.138\text{eV}}\\ =1090\text{nm}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $-2.04\text{eV}$ for $E_2$, and $-0.508\text{eV}$ for $E_4$ in the equation (VII) to find ${\lambda }_{\beta }$

  $\begin{array}{c}{\lambda }_{\beta }=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(-0.508\text{eV}\right)-\left(-2.04\text{eV}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(1.532\text{eV}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{1.532\text{eV}}\\ =811\text{nm}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $-2.04\text{eV}$ for $E_2$, and $-0.325\text{eV}$ for $E_5$ in the equation (VIII) to find ${\lambda }_{\gamma }$

  $\begin{array}{c}{\lambda }_{\gamma }=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(-0.325\text{eV}\right)-\left(-2.04\text{eV}\right)}\\ =\frac{19.878\times {10}^{-26}\text{J}\cdot \text{m}\left(\frac{6.242\times {10}^{18}\text{eV}}{1\text{J}}\right)}{\left(1.715\text{eV}\right)}\\ =\frac{1240\text{eV}\cdot \text{nm}}{1.715\text{eV}}\\ =724\text{nm}\end{array}$

Similarly the shortest wavelength is, $609\text{nm}$.

Therefore, the wavelength of the first three lines is ${\lambda }_{\alpha }=1090\text{nm}$, ${\lambda }_{\beta }=811\text{nm}$, ${\lambda }_{\alpha }=724\text{nm}$ and one short wavelength limit in Balmer series is $609\text{nm}$.

(c)

To determine

Show that the wavelength of the first four lines and short wavelength limit of the Lyman series.

Answer to Problem 45P

The wavelength of the first four lines is $122\text{nm, 103}\text{nm, 97}\text{.3}\text{nm, 95}\text{.0}\text{nm}$ and short wavelength limit of the Lyman series is $91.2\text{nm}$.

Explanation of Solution

Write the expression for wavelength of the Lyman series for hydrogen atom.

  $\frac{1}{\lambda }=R_H\left(1-\frac{1}{n^2}\right)$        (X)

Here, $R_H$ is the Rydberg constant and $n$ is the energy level.

Conclusion:

For the $\alpha$ spectral line, substitute $2$ for $n$, $1.097\times {10}^7/\text{m}$ for $R_H$, and ${\lambda }_{\alpha }$ for $\lambda$ in the equation (X).

  $\begin{array}{c}\frac{1}{{\lambda }_{\alpha }}=\left(1.097\times {10}^7/\text{m}\right)\left(1-\frac{1}{2^2}\right)\\ {\lambda }_{\alpha }=122\text{nm}\end{array}$

For the $\beta$ spectral line, substitute $3$ for $n$, $1.097\times {10}^7/\text{m}$ for $R_H$, and ${\lambda }_{\beta }$ for $\lambda$ in the equation (X).

  $\begin{array}{c}\frac{1}{{\lambda }_{\beta }}=\left(1.097\times {10}^7/\text{m}\right)\left(1-\frac{1}{3^2}\right)\\ {\lambda }_{\beta }=103\text{nm}\end{array}$

For the $\gamma$ spectral line, substitute $4$ for $n$, $1.097\times {10}^7/\text{m}$ for $R_H$, and ${\lambda }_{\gamma }$ for $\lambda$ in the equation (X).

  $\begin{array}{c}\frac{1}{{\lambda }_{\gamma }}=\left(1.097\times {10}^7/\text{m}\right)\left(1-\frac{1}{4^2}\right)\\ {\lambda }_{\gamma }=97.2\text{nm}\end{array}$

For the $\delta$ spectral line, substitute $5$ for $n$, $1.097\times {10}^7/\text{m}$ for $R_H$, and ${\lambda }_{\delta }$ for $\lambda$ in the equation (X).

  $\begin{array}{c}\frac{1}{{\lambda }_{\delta }}=\left(1.097\times {10}^7/\text{m}\right)\left(1-\frac{1}{5^2}\right)\\ {\lambda }_{\gamma }=94.9\text{nm}\end{array}$

For the short wavelength, substitute $\infty$ for $n$, $1.097\times {10}^7/\text{m}$ for $R_H$, and ${\lambda }_{\text{short}}$ for $\lambda$ in the equation (X).

  $\begin{array}{c}\frac{1}{{\lambda }_{\text{short}}}=\left(1.097\times {10}^7/\text{m}\right)\left(1-\frac{1}{{\infty }^2}\right)\\ {\lambda }_{\text{short}}=91.1\text{nm}\end{array}$

Therefore, the wavelength of the first four lines is $122\text{nm, 103}\text{nm, 97}\text{.3}\text{nm, 95}\text{.0}\text{nm}$ and short wavelength limit of the Lyman series is $91.2\text{nm}$.

(d)

To determine

From the observation, explain this atom as an hydrogen.

Answer to Problem 45P

The spectrum series of the atom proved that this is hydrogen.

Explanation of Solution

Write the expression for Doppler shift.

  $\begin{array}{c}\frac{f^{\prime }}{f}=\frac{{\lambda }^{\prime }}{\lambda }\\ =\sqrt{\frac{c-v}{c+v}}\end{array}$        (XI)

Here, $f$ is the frequency of the observer, $f$ is the actual frequency, $\lambda$ is the wavelength of the shift, ${\lambda }^{\prime }$ is the wavelength of the source moving away, and $v$ is the velocity of the source.

Conclusion:

The required speed of the source is $v=0.471c$.

The spectrum could be that of hydrogen, Doppler shifted by motion away from us at speed $0.471c$