Q48P - Fundamentals Of Physics

Question

Light of wavelength 121.6 nm is emitted by a hydrogen atom. What are the (a) higher quantum number and (b) lower quantum number of the transition producing this emission? (c) What is the series that includes the transition? 

Step-by-Step Solution

Answer

(a) Higher quantum number, ${\mathrm{n}}_1=2$ 

(b) Lower quantum  number, ${\mathrm{n}}_2=1$

(c) Lyman series.

Step 1: Identification of the given data:

The given data is listed below.

Wavelength of light is $\mathrm{\lambda }=121.6 \mathrm{nm}$.

Step 2: The principal Quantum number:

The principal quantum number is used to describe the electron’s state and is the one-four quantum number assigned to each electron in an atom.

The value of the principal quantum number is a natural number.

Step 3: (a) Determine the value of the higher quantum number:

The energy emitted by the hydrogen atom of the photon is expressed as,

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

Now, multiply both sides by $\frac{1}{hc}$, and you have

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

Also as the wavelength is,

$\frac{1}{\mathrm{\lambda }}=\frac{{\mathrm{E}}_{\mathrm{Ph}}}{\mathrm{hc}}$ 

Substitute$\frac{A}{hc}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$for$\frac{{\mathrm{E}}_{\mathrm{P}}\mathrm{h}}{\mathrm{hc}}$ in the above equation.

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

 

Substitute 121.6 nm for $\lambda$, $6.26\times {10}^{-34} \mathrm{J}.\mathrm{s}$ for h , $3\times {10}^8 \mathrm{m}/\mathrm{s}$ for c , and 13.6 eV for A in the above equation.

$$\begin{gathered} \frac{\left(6.26\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3\times {10}^8 \mathrm{m}/\mathrm{s}\right)}{\left(121.6\times {10}^{-9} \mathrm{m}\right)\left(13.6\times 1.602\times {10}^{-9} \mathrm{J}\right)}=\left(\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}\right) \\ 0.75=\left(\frac{1}{{\mathrm{n}}_2^2}-\frac{1}{{\mathrm{n}}_1^2}\right) \end{gathered}$$ 

To satisfy the above equation, the values of ${\mathrm{n}}_2$ and ${\mathrm{n}}_1$ are 1 and 2 respectively. Thus,

$\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}=0.75$ 

Thus, the higher quantum number is  ${\mathrm{n}}_1=2$.

Step 4: (b) Determine the value of lower quantum number:

And so,

$\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}=0.75$ 

Thus, the lower quantum number is $n_2=1$.

Step 5: (c) Determine the name of the series that includes the transition:

The name of the series that includes the transition is Lyman series transition as this transition occurs when the electron goes from $n_1$ to $n_2$ emitting a photon.