Question

What are the (a) wavelength range and (b) frequency range of the Lyman series? What are the (c) wavelength range and (d) frequency range of the Balmer series?

Short answer

(a) The wavelength range is 30.4.

(b) The frequency range of the Lyman series is $8.22\times {10}^{14}\mathrm{Hz}$8.

(c) The wavelength range is 292 nm.

(d) The frequency range of the Balmar series is $3.65\times {10}^{14}\mathrm{Hz}$.

Step-by-step solution

The energy of the photon emitted by a hydrogen atom:

The expression of the energy of the photon emitted for a hydrogen atom jumps from a state of ${\mathbf{n}}_{\mathbf{1}}$to ${\mathbf{n}}_{\mathbf{2}}$is given by,

$\mathbf{E}\mathbf{=}\mathbf{A}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Here, the state ${\mathbf{n}}_{\mathbf{2}}\mathbf{>}{\mathbf{n}}_{\mathbf{1}}$, and A=13.6 eV .

Multiply both sides with $\frac{\mathbf{1}}{\mathbf{hc}}$.

$\frac{\mathbf{E}}{\mathbf{hc}}\mathbf{=}\frac{\mathbf{A}}{\mathbf{hc}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

It is known that

$\frac{\mathbf{1}}{\mathbf{\lambda }}\mathbf{=}\frac{\mathbf{E}}{\mathbf{hc}}$

$\frac{\mathbf{1}}{\mathbf{\lambda }}\mathbf{=}\frac{\mathbf{A}}{\mathbf{hc}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

(a) Define the wavelength range Lyman series:

The Lyman series is associated with transitions to or from the n=1 level of the hydrogen atom, the shortest wavelength occur between $n_1=1$and $n_2=\infty$.

The wavelength is,

$$\begin{gathered} \frac{1}{{\mathrm{\lambda }}_{\mathrm{L},\mathrm{short}}}=\frac{\mathrm{A}}{\mathrm{hc}}\left(1-\frac{1}{\infty }\right) \\ =\frac{\mathrm{A}}{\mathrm{hc}} \\ {\mathrm{\lambda }}_{\mathrm{L},\mathrm{short}}=\frac{\mathrm{hc}}{\mathrm{A}} \end{gathered}$$

The longest wavelength occur between $n_1=1$and ${\mathrm{n}}_2=2$.

The wavelength is,

$$\begin{gathered} \frac{1}{{\lambda }_{L,long}}=\frac{A}{hc}\left(1-\frac{1}{4}\right) \\ =\frac{3A}{4hc} \\ {\lambda }_{L,long}=\frac{4hc}{A} \end{gathered}$$

The range of the wavelength is just the difference between the longest and shortest wavelengths.

$$\begin{gathered} ∆{\lambda }_1={\lambda }_{L,long}-{\lambda }_{L,short} \\ =\frac{4hc}{3A}-\frac{hc}{A} \\ ∆{\lambda }_L=\frac{hc}{3A} \end{gathered}$$

….. (1)

Substitute $4.136\times {10}^{-15}\mathrm{eV}.\mathrm{s}$for h, $3\times {10}^8\mathrm{m}/\mathrm{s}$for c, and 13.6 eV for A in equation (1).

$$\begin{gathered} ∆{\mathrm{\lambda }}_{\mathrm{L}}=\frac{\left(4.136\times {10}^{-15}\mathrm{eV}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{3\left(13.6\mathrm{eV}\right)} \\ =30.4\times {10}^{-9}\mathrm{m} \\ =30.4\mathrm{nm} \end{gathered}$$

Therefore, the wavelength range is 30.4 nm.

(b) Find the frequency range of the Lyman series:

The frequency is the speed of light divided by the wavelength.

$$\begin{gathered} ∆f_L=\frac{c}{{\lambda }_{L,short}}-\frac{c}{{\lambda }_{L,long}} \\ =\frac{A}{h}-\frac{3A}{h} \end{gathered}$$

$∆f_L=\frac{A}{4h}$ ….. (2)

Substitute $4.136\times {10}^{-15}\mathrm{eV}.\mathrm{s}$for h, and 13.6 eV for A in equation (2).

$$\begin{gathered} ∆{\mathrm{f}}_{\mathrm{L}}=\frac{13.6\mathrm{eV}}{4\left(4.136\times {10}^{-15}\mathrm{eV}.\mathrm{s}\right)} \\ =8.22\times {10}^{14}\mathrm{Hz} \end{gathered}$$

Hence, the frequency range of the Lyman series is $8.22\times {10}^{14}\mathrm{Hz}$.

(c) Determine the wavelength range of the Balmer series:

The Balmer series is associated with transitions to or from the n=2 level of the hydrogen atom, the shortest wavelength occur between $n_1=2$and $n_2=\infty$.

The wavelength is,

$$\begin{gathered} \frac{1}{{\lambda }_{B,short}}=\frac{A}{hc}\left(\frac{1}{4}-\frac{1}{\infty }\right) \\ =\frac{A}{4hc} \\ {\lambda }_{B,short}=\frac{4hc}{A} \end{gathered}$$

The longest wavelength occur between $n_1=2$and $n_2=3$.

The wavelength is,

$$\begin{gathered} \frac{1}{{\lambda }_{B,long}}=\frac{A}{hc}\left(\frac{1}{4}-\frac{1}{9}\right) \\ =\frac{5A}{36hc} \\ {\lambda }_{B,long}=\frac{36hc}{5A} \end{gathered}$$

The range of the wavelength is just the difference between the longest and shortest wavelengths.

$$\begin{gathered} ∆{\lambda }_B={\lambda }_{B,long}-{\lambda }_{B,short} \\ =\frac{36hc}{5A}-\frac{4hc}{A} \\ ∆{\lambda }_B=\frac{16hc}{5A} \end{gathered}$$

….. (3)

Substitute $4.136\times {10}^{-15}eV\cdot s$for h,$3\times {10}^8\hspace{0.33em}\frac{m}{s}$for c , and 13.6 eV for A in equation (1).

$$\begin{gathered} ∆{\mathrm{\lambda }}_{\mathrm{B}}=\frac{16\left(4.136\times {10}^{-15}\mathrm{eV}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{5\left(13.6\mathrm{eV}\right)} \\ =2.92\times {10}^{-7}\mathrm{m} \\ =292\mathrm{mm} \end{gathered}$$

Therefore, the wavelength range is 292 nm.

(d) Define the frequency range of the Balmar series:

The frequency is the speed of light divided by the wavelength.

$$\begin{gathered} ∆f_B=\frac{c}{{\lambda }_{B,short}}-\frac{c}{{\lambda }_{B,long}} \\ =\frac{1A}{4h}-\frac{5A}{36h} \end{gathered}$$

$∆f_B=\frac{A}{9h}$ ….. (4)

Substitute $4.136\times {10}^{-15}eV\cdot s$for h , and 13.6 eV for A in equation (2).

$$\begin{gathered} ∆{\mathrm{f}}_{\mathrm{B}}=\frac{13.6\mathrm{eV}}{9\left(4.136\times {10}^{-15}\mathrm{eV}.\mathrm{s}\right)} \\ =3.65\times {10}^{14}\mathrm{Hz} \end{gathered}$$

Hence, the frequency range of the Balmar series is $3.65\times {10}^{14}\mathrm{Hz}$.