Question

Calculate the wavelength of radiation emitted when an electron in a hydrogen atom jumps from $n = \infty $ to $n = 1$.

Answer 1

Hint The transition is from $n = \infty $ to $n = 1$ to Lyman’s series. The higher the difference between the levels of transition, the higher the energy of the radiation released and thus, the lower the wavelength.

Formula used:
$\Rightarrow \dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right)$ where $\lambda $ is the wavelength of the radiation, ${n_f}$ is the final orbit, ${n_i}$ is initial orbit and ${R_H}$ is the Rydberg constant.

Complete step by step answer
Whenever an electron in a hydrogen atom transitions from $n > 1$ to $n = 1$ (where $n$ is the principal quantum number taking the values 1, 2, 3….), it is called the Lyman series. Other series such as Balmer, Paschen, etc which all have their respective value of $n$. The Balmer series, for example, is a transition from $n > 2$ to $n = 2$.
Whenever an electron transitions from a higher orbit to a lower orbit light is emitted by the electron and the wavelength of this light emitted is given by the Rydberg formula
$\Rightarrow \dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right)$
Where $\lambda $ is the wavelength of the radiation, ${n_f}$ is the final orbit, ${n_i}$ is initial orbit and ${R_H}$ is the Rydberg constant which is equal to $1.097 \times {10^7}{m^{ - 1}}$. For Lyman’s series, ${n_f} = 1$.
Therefore, we have
$\Rightarrow \dfrac{1}{\lambda } = {R_H}\left( {1 - \dfrac{1}{{n_i^2}}} \right)$
In the question, ${n_i} = \infty $.
Hence, substituting the values and solving we get,
$\Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}{m^{ - 1}}\left( {1 - \dfrac{1}{\infty }} \right) $
$\Rightarrow 1.097 \times {10^7}{m^{ - 1}} $
Inverting both sides, we get
$\Rightarrow \lambda = \dfrac{1}{{1.097 \times {{10}^7}}} $
$\Rightarrow 9.12 \times {10^{ - 8}}m $
$\therefore \lambda = 9.12 \times {10^{ - 8}}m$

Note
An alternate method is to use the formula
$\Rightarrow {E_n} = - 2.178 \times {10^{ - 18}}\left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right)$
This allows us to calculate the energy released as follows
$\Rightarrow {E_n} = - 2.178 \times {10^{ - 18}}\left( {\dfrac{1}{1} - \dfrac{1}{\infty }} \right) $
$\Rightarrow - 2.178 \times {10^{ - 18}}J $
The negative sign implies that energy was released during the transition and can be neglected in our quest to find the wavelength. Then we can recall that the energy is related to the wavelength by
$\Rightarrow E = \dfrac{{hc}}{\lambda }$
Where $c$ is the speed of light which is given by $c \approx 3 \times {10^8}m/s$ and $h$ is called the Planck’s constant which is given by $h \approx 6.63 \times {10^{ - 34}}{m^2}kg/s$.
Rearranging the formula $E = \dfrac{{hc}}{\lambda }$ we get
$\Rightarrow \lambda = \dfrac{{hc}}{E}$.