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Question: The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to sta...

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be: [Rydberg constant=1.097×107 m−11.097 \times {10}^7 \,{m^{ - 1}}]
A. 9191nm
B. 192192 nm
C. 406406 nm
D. 9.1×10−89.1 \times {10^{ - 8}} m

Explanation

Solution

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, Ef{E_f} is its final energy after the transition, and the photon’s energy is hv=hcλhv = \dfrac{{hc}}{\lambda }, then conservation of energy gives,
hv=hcλ=Ei−Efhv = \dfrac{{hc}}{\lambda } = {E_i} - {E_f}
Now we can use the formula En=−Rhcn2,n=1,2,3,...{E_n} = - \dfrac{{Rhc}}{{{n^2}}},n = 1,2,3,...
So we get final formula as:
ΔE =En2−En1 ΔE =−Rhcn22−(−Rhcn12)\Delta E \,= E_{n_2}-E_{n_1} \\\ \Delta E \, = - \dfrac{{Rhc}}{{{{n_2}^2}}} - (- \dfrac{{Rhc}}{{{{n_1}^2}}})
hcλ\dfrac{{hc}}{\lambda } = −Rhcn22−(−Rhcn12)- \dfrac{{Rhc}}{{{{n_2}^2}}} - (- \dfrac{{Rhc}}{{{{n_1}^2}}})
⇒1λ=R(1n12−1n22)\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{{n_2} ^2}}}} \right)
Here in this question we have n1=∞{n_1} = \infty and n2=1{n_2} = 1so let’s put this data into formula,
⇒1λ=R(112−1∞2)\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{1_{}}^2}} - \dfrac{1}{{{\infty ^2}}}} \right)
⇒1λ=R\Rightarrow \dfrac{1}{\lambda } = R
⇒1R=λ\Rightarrow \dfrac{1}{R} = \lambda
By putting R= 1.097×107 m−11.097 \times {10}^7 \, {m^{ - 1}}
We get λ\lambda = 9.1×10−89.1 \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 Hα{H_\alpha } put n=3 and for Hβ{H_\beta } put n=4. Put n1{n_1} and n2{n_2} values very carefully because that is the most important part and remember Rydberg constant value.