Question

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be: [Rydberg constant=$1.097 \times {10}^7 \,{m^{ - 1}}$]
A. $91$nm
B. $192$ nm
C. $406$ nm
D. $9.1 \times {10^{ - 8}}$ m

Answer 1

Hint: We can solve this question by Bohr transition theory since there is a transition from infinity to state 1 we can apply conservation of energy and we can calculate the final wavelength.

Complete step by step solution:
Let us first understand what Bohr concludes about transition.
According to Bohr when an atom makes a transition from higher energy level to lower level it emits a photon with energy equal to the energy difference between the initial and final levels. If is the initial energy of the atom before such as a transition, ${E_f}$ is its final energy after the transition, and the photon’s energy is $hv = \dfrac{{hc}}{\lambda }$, then conservation of energy gives,
$hv = \dfrac{{hc}}{\lambda } = {E_i} - {E_f}$
Now we can use the formula ${E_n} = - \dfrac{{Rhc}}{{{n^2}}},n = 1,2,3,...$
So we get final formula as:
$\Delta E \,= E_{n_2}-E_{n_1} \\
 \Delta E \, = - \dfrac{{Rhc}}{{{{n_2}^2}}} - (- \dfrac{{Rhc}}{{{{n_1}^2}}}) $
$ \dfrac{{hc}}{\lambda }$ = $ - \dfrac{{Rhc}}{{{{n_2}^2}}} - (- \dfrac{{Rhc}}{{{{n_1}^2}}}) $
\[\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{{n_2} ^2}}}} \right)\]
Here in this question we have ${n_1} = \infty $ and ${n_2} = 1$so let’s put this data into formula,
\[\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1_{}}^2}} - \dfrac{1}{{{\infty ^2}}}} \right)\]
\[\Rightarrow \dfrac{1}{\lambda } = R\]
\[\Rightarrow \dfrac{1}{R} = \lambda \]
By putting R= $1.097 \times {10}^7 \, {m^{ - 1}}$
We get $\lambda$ = $9.1 \times {10^{ - 8}}$ m

So our correct option is D

Note:
Always use this formula when there is a transition from higher state to lower state .One more thing is that for \[{H_\alpha }\] put n=3 and for \[{H_\beta }\] put n=4. Put \[{n_1}\] and \[{n_2}\] values very carefully because that is the most important part and remember Rydberg constant value.