Question

In the Balmer series for hydrogen atoms, find the energy of the photon corresponding to the longest wavelength.
A.$18.9\;{\rm{eV}}$
B.$3.03\;{\rm{eV}}$
C.$1.89\;{\rm{eV}}$
D.$30.3\;{\rm{eV}}$

Answer 1

Hint: To find the energy of photons for the longest wavelength, firstly, we will find the length of the maximum wavelength between the principal quantum number and the consecutive quantum number. We will use the Rydberg equation to express the relation between maximum wavelength, consecutive quantum numbers, and the atomic number. We will substitute the values of consecutive quantum numbers, Rydberg's constant, and the atomic number of hydrogen to find the maximum wavelength. Then we will express the relation for the energy of photons in terms of Planck's constant and frequency. We can also write the frequency as the ratio of the speed of light and the wavelength of the spectral lines. We will then substitute the speed of light, maximum wavelength, and Planck's constant to find the energy.

Formula used: We will use the Rydberg equation which can be expressed as
$\dfrac{1}{\lambda } = R{z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where $R$ is Rydberg's constant, ${n_2}$ and ${n_1}$ are the quantum numbers, $\lambda $ is the wavelength and $z$ is the atomic number.
We also use the expression for energy of photons.
$E = h\upsilon $
Where $h$ is the planck’s constant and $\upsilon $ represents the frequency.

Complete step by step answer:
Given: The given atom is hydrogen.

The longest wavelength are found in Balmer series when a transition undergoes from ${n_2} = 3$ and ${n_1} = 2$
We will use the Rydberg equation to express the relation between maximum wavelength $\lambda $ , quantum numbers $n$.
$\dfrac{1}{\lambda } = R{z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right)$
Where $R$ is the Rydberg constant such that its value for hydrogen is $1.097 \times {10^7}\;{\rm{per}}\;{\rm{metre}}$and $z$ is the atomic number. For hydrogen, we have an atomic number as 1.
We will substitute 3 for ${n_2}$ and 2 for ${n_1}$ in the above expression.
\[\begin{array}{l}
\lambda = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)\\
\lambda = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\dfrac{5}{{36}}} \right)\\
\lambda = 6.564 \times {10^{ - 7}}\;{{\rm{m}}^{ - 1}}
\end{array}\] ……(i)

Now, we write the expression for the energy of photons which can be expressed as
$E = h\upsilon $ ……(ii)
Where $h$ is the planck’s constant and its value is $6.6 \times {10^{ - 34}}\;{\rm{Js}}$ and $\upsilon $ represents the frequency.
We also know that the frequency can be expressed in the term of wavelength $\lambda $ and speed of light $c$ .
$\upsilon = \dfrac{c}{\lambda }$ ……(iii)
We will substitute the values of $\upsilon $ from equation (iii) in equation (i) which can be expressed as
$E = h\dfrac{c}{\lambda }$
We will substitute $6.6 \times {10^{ - 34}}\;{\rm{Js}}$for $h$,\[6.564 \times {10^{ - 7}}\;{{\rm{m}}^{ - 1}}\] for $\lambda $ and $3.00 \times {10^8}\;{\rm{m}}{{\rm{s}}^{ - 1}}$ for $c$ in the above expression.
$\begin{array}{l}
E = \dfrac{{\left( {6.6 \times {{10}^{ - 34}}\;{\rm{Js}}} \right) \times \left( {3.00 \times {{10}^8}\;{\rm{m}}{{\rm{s}}^{ - 1}}} \right)}}{{\left( {6.564 \times {{10}^{ - 7}}\;{\rm{m}}} \right)}}\\
E = 3.02 \times {10^{ - 19}}\;{\rm{J}}
\end{array}$
We know that $1\;{\rm{Joule}}$ is equivalent to $6.242 \times {10^{18}}\;{\rm{eV}}$ . Hence to convert the energy found in ${\rm{Joule}}$ we will multiply the result with to $6.242 \times {10^{18}}\;{\rm{eV}}$which can be expressed as
$\begin{array}{l}
E = 3.02 \times {10^{ - 19}} \times 6.242 \times {10^{18}}\;{\rm{eV}}\\
E = 1.89\;{\rm{eV}}
\end{array}$
Therefore, the energy of photon corresponding to longest wavelength is $1.89\;{\rm{eV}}$

So, the correct answer is “Option C”.

Note:
The energy that we will get from solving the equation will be in Joule. Hence we need to convert the energy in Joule into the energy in electron Volts. In this question we were asked to find the energy of photons for maximum wavelength. But, if any question the energy of photons for minimum wavelength is asked then we can consider transition between ${n_2} = \infty $ and ${n_1} = 2$ .