Question

When the system de-excites from its excited state to the ground state, the wavelength of radiation is

(A) $1217A^\circ$

(B) $2431A^\circ$

(C) $608A^\circ$

(D) None of these

Answer 1

The given problem can be solved by taking the consideration of Bohr’s atomic model. As we know that in Bohr’s atomic model of hydrogen, when an atom of hydrogen receives energy by some processes such as electron collision or heat, then the atom may get sufficient energy to raise the electron from a lower energy state to a higher energy state.

Then this atom is said to be in an excited state.

Complete step by step answer:
Step 1:

If one system is kept at room temperature then atoms of that system are in ground state or lower energy state. If the atom of this system gets energy by some means such as collision or electrons or heat, then these atoms may get sufficient energy to excite the electron to higher energy levels i.e., from $\mathop n\nolimits_1 = 1$ (ground energy state) to $\mathop n\nolimits_2 = 2,3,4, - - - - -$ (higher energy state).

The atom is then said to be in the excited state. But an electron that is excited to higher energy levels is not stable at those levels so it falls back to the state of lower energy emitting a photon of particular energy and the energy of this emitted photon is equal to the difference in the energy of two states.

Let $\mathop E\nolimits_1 =$ total energy of electron in the inner orbit ($\mathop n\nolimits_1$ th) i.e. ground (stable) state

$\mathop E\nolimits_2 =$ total energy of electron in the outer orbit ($\mathop n\nolimits_2$ th) i.e. excited (unstable) state

When an electron jumps from an outer to inner orbit, the energy of emitted photon is given by –

$h\nu = \mathop E\nolimits_2 - \mathop E\nolimits_1$

Step 2: For the spectrum of the system, wavelength ($\lambda$) or wavenumber ($\overline \nu$) can be given by the ‘Rydberg formula for the spectrum of the system’ and defined as given below –

$\dfrac{1}{\lambda } = \overline \nu = R\left( {\dfrac{1}{{\mathop n\nolimits_1^2 }} - \frac{1}{{\mathop n\nolimits_2^2 }}} \right)$ (1)

Where $\mathop n\nolimits_1 = 1$ inner orbit i.e. ground state

$\mathop n\nolimits_2 = 2$ outer orbit i.e. excited state

And $R = 1.097 \times \mathop {10}\nolimits^7 m^{-1}$ (Rydberg constant)

Now from equation (1)

$\dfrac{1}{\lambda } = 1.097 \times \mathop {10}\nolimits^7 \times \dfrac{3}{4}$

$\dfrac{1}{\lambda } = 0.82275 \times \mathop {10}\nolimits^7$; simplifying the above equation

$\lambda = \dfrac{1}{{0.82275 \times \mathop {10}\nolimits^7 }}$

$\lambda = 1.215 \times \mathop {10}\nolimits^{ - 7} m = 1215{\rm A}^\circ$

$\lambda = 1215{\rm A}^\circ \approx 1217{\rm A}^\circ$

$\therefore$ Hence, option (A) is correct.

Note:

(i) As in the final answer instead of $\lambda = 1215{\rm A}^\circ$, $\lambda \simeq 1217{\rm A}^\circ$ is taken into account because this wavelength is at very small at the scale of $\mathop {10}\nolimits^{-10}$.

(ii) From equation (1) it is clear that wavelengths/frequencies/wavenumbers of radiations emitted by the excited system are not continuous. They are having specific values depending upon the values of $\mathop n\nolimits_1$ and $\mathop n\nolimits_2$.