Question

Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2.
(a) Find the smallest wavelength emitted by hydrogen.
(b) List the wavelengths emitted by hydrogen in the visible range (380 nm to 780 nm)

Answer 1

Hint:The energy difference for the transition in energy state is larges in the minimum orbit transition. The energy of the wave is inversely proportional to the wavelength.

Formula used:
\[E = \dfrac{{hc}}{\lambda }\]
where E is the energy of wave with wavelength \[\lambda \]
\[\Delta E = \left( { - 13.6eV} \right)\left( {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_1^2}}} \right)\]
where \[\Delta E\] is the energy difference for the transition between the energy level \[{n_1}\] to \[{n_2}\].

Complete step by step solution:
(a) The energy difference between two principal quantum number is given as,
\[\Delta E = \left( { - 13.6eV} \right)\left( {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_1^2}}} \right)\]
Here, \[{n_1}\] is the initial principal quantum number and \[{n_2}\]is the final quantum number. Here, the change in the principal quantum number is 2.

If the initial principal quantum number is \[n\] then the final principle quantum number is \[n - 2\]. So, the energy difference will be,
\[\Delta E = \left( { - 13.6eV} \right)\left( {\dfrac{1}{{{{\left( {n - 2} \right)}^2}}} - \dfrac{1}{{{n^2}}}} \right) \\ \]
As we know that the energy is inversely proportional to the wavelength. So, for the minimum wavelength the energy difference should be maximum. The maximum energy difference is between the ground state and other states. For the ground state, the other state should be 3 as the change is given as 2. So,
\[{n_2} = 3\]and \[{n_1} = 1\]
Putting the value in the given expression for the energy difference, we get
\[\Delta E = \left( { - 13.6eV} \right)\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{1^2}}}} \right) \\ \]
\[\Rightarrow \Delta E = 12.1\,eV\]
Using the energy formula,
\[E = \dfrac{{hc}}{\lambda } \\ \]
\[\Rightarrow \lambda = \dfrac{{hc}}{E}\]
Putting the values, we get
\[\lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{12.1 \times 1.6 \times {{10}^{ - 19}}}}m \\ \]
\[\Rightarrow \lambda = 1.03 \times {10^{ - 7}}m \\ \]
\[\therefore \lambda = 103\,nm\]
Hence, the required minimum wavelength is 103 nm.

Therefore, the correct answer is 103 nm.

(b) Let the transition take place from the first excited state. As the change is given as 2, so the next energy state will be 4.
\[{n_2} = 4\]and \[{n_1} = 2\]
Putting the value in the given expression for the energy difference, we get
\[\Delta E = \left( { - 13.6eV} \right)\left( {\dfrac{1}{{{4^2}}} - \dfrac{1}{{{2^2}}}} \right) \\ \]
\[\Rightarrow \Delta E = 2.55\,eV\]
Using the energy formula,
\[E = \dfrac{{hc}}{\lambda } \\ \]
\[\Rightarrow \lambda = \dfrac{{hc}}{E}\]
Putting the values, we get
\[\lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2.55 \times 1.6 \times {{10}^{ - 19}}}}m \\ \]
\[\Rightarrow \lambda = 4.87 \times {10^{ - 7}}m \\ \]
\[\therefore \lambda = 487\,nm\]
Hence, from the given range of wavelength the obtained wavelength is 487 nm.

Therefore, the correct answer is 487 nm.

Note: The largest energy of transition is the ionization of the electron from the ground state. As the change in the principal quantum number is restricted we cannot use the ionization condition.