Chapter 18, Problem 2SP

(a)

To determine

The transition in the Balmer series which produces the smallest frequency photon.

Answer to Problem 2SP

The transition in the Balmer series which produces the smallest frequency photon is $m=3$ to $n=2$.

Explanation of Solution

Hydrogen spectrum lies in the visible, ultraviolet as well as infrared regions. Hydrogen spectrum is divided into different spectral series named as Lyman series, Balmer series, Paschen series, Brackett series and Pfund series. The wavelengths of lines in each series is determined using Balmer’s formula.

The Balmer’s formula is given as $\frac{1}{\lambda }=R\left(\frac{1}{n^2}-\frac{1}{m^2}\right)$ where $\lambda$ is the wavelength of the line, $n$ is an integer which represents a particular series or in other words $n$ will be a constant for a particular series, $m$ is another integer which will be always greater than $n$ and $R$ is the Rydberg constant. The value of $n$ for Balmer series is $2$.

From the Balmer’s formula it is clear that the photon having smallest frequency or longest wavelength will be emitted when the value of $m$ is greater than $n$ by just $1$. This implies the value of $m$ to produce the smallest frequency photon in the Balmer series should be $3$.

Conclusion:

Thus the $m=3$ to $n=2$ transition produces the smallest frequency photon in the Balmer series.

(b)

To determine

The energy difference in joules for the two levels involved in the transition of part (a).

Answer to Problem 2SP

The energy difference in joules for the two levels involved in the transition of part (a) is $3.02\times {10}^{-19}\text{ J}$.

Explanation of Solution

Write the Balmer’s formula.

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

The value of $R$ is $1.097\times {10}^7{\text{ m}}^{-1}$.

Substitute $1.097\times {10}^7{\text{ m}}^{-1}$ for $R$,$2$ for $n$ and $3$ for $m$ in equation (1) to find $\left(1/\lambda \right)$.

$\begin{array}{c}\frac{1}{\lambda }=1.097\times {10}^7{\text{ m}}^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\\ =1523611.1{\text{ m}}^{-1}\end{array}$

Write the equation for the energy of a photon.

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

Here,

$E$ is the energy of the photon

$h$ is the Planck’s constant

$c$ is the speed of light in vacuum

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 $1523611.1{\text{ m}}^{-1}$ for $\left(1/\lambda \right)$ in the above equation to find $E$.

$\begin{array}{c}E=\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{ m/s}\right)\left(1523611.1{\text{ m}}^{-1}\right)\\ =3.02\times {10}^{-19}\text{ J}\end{array}$

The energy of the emitted photon is equal to the energy difference between the levels involved in the transition.

Conclusion:

Thus the energy difference in joules for the two levels involved in the transition of part (a) is $3.02\times {10}^{-19}\text{ J}$.

(c)

To determine

The frequency and wavelength of the photon emitted in the transition.

Answer to Problem 2SP

The frequency of the photon emitted in the transition is $4.56\times {10}^{14}\text{ Hz}$ and the wavelength is $657\text{ nm}$.

Explanation of Solution

From part (b),

$\frac{1}{\lambda }=1523611.1{\text{ m}}^{-1}$

This gives,

$\begin{array}{c}\lambda =\frac{1}{1523611.1{\text{ m}}^{-1}}\\ =657\times {10}^{-9}\text{ m}\\ =657\text{ nm}\end{array}$

Write the equation for the frequency of the photon.

$f=\frac{c}{\lambda }$ (2)

Here,

$f$ is the frequency of the photon

Substitute $3.0\times {10}^8\text{ m/s}$ for $c$ and $657\times {10}^{-9}\text{ m}$ for $\lambda$ in equation (2) to find $f$.

$\begin{array}{c}f=\frac{3.0\times {10}^8\text{ m/s}}{657\times {10}^{-9}\text{ m}}\\ =4.56\times {10}^{14}\text{ Hz}\end{array}$

Conclusion:

Thus the frequency of the photon emitted in the transition is $4.56\times {10}^{14}\text{ Hz}$ and the wavelength is $657\text{ nm}$.

(d)

To determine

The frequency and wavelength of the photon with the longest wavelength in the Lyman series.

Answer to Problem 2SP

The frequency of the photon with the longest wavelength in the Lyman series is $2.46\times {10}^{15}\text{ Hz}$ and its wavelength is $122\text{ nm}$.

Explanation of Solution

The value of $n$ for Lyman series is $1$ so that the longest wavelength photon is emitted for the $m=2$ to $n=1$ transition.

Substitute $1.097\times {10}^7{\text{ m}}^{-1}$ for $R$,$1$ for $n$ and $2$ for $m$ in equation (1) to find $\left(1/\lambda \right)$.

$\begin{array}{c}\frac{1}{\lambda }=1.097\times {10}^7{\text{ m}}^{-1}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\\ =8227500{\text{ m}}^{-1}\end{array}$

This gives,

$\begin{array}{c}\lambda =\frac{1}{8227500{\text{ m}}^{-1}}\\ =122\times {10}^{-9}\text{ m}\\ =122\text{ nm}\end{array}$

Substitute $3.0\times {10}^8\text{ m/s}$ for $c$ and $122\times {10}^{-9}\text{ m}$ for $\lambda$ in equation (2) to find $f$.

$\begin{array}{c}f=\frac{3.0\times {10}^8\text{ m/s}}{122\times {10}^{-9}\text{ m}}\\ =2.46\times {10}^{15}\text{ Hz}\end{array}$

Conclusion:

Thus the frequency of the photon with the longest wavelength in the Lyman series is $2.46\times {10}^{15}\text{ Hz}$ and its wavelength is $122\text{ nm}$.