Which of the following transition will have minimum wavelength and why n4-n1 n4-n2 n2-n1

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`n_4 to n_1``n_2 to n_1``n_4 to n_2``n_3 to n_1`

Answer : A

Solution : For `n_4 to n_1` , greater transition, greater the energy difference, lesser will be the wavelength .

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Hint: Think about the particles present in an atom and the forces present in an atom. In the question, n is the principal quantum number. We have the formula for calculating wavelength of a photon as $\dfrac{1}{\lambda }=R\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)$. Wavelength of photon is inversely proportional to $\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)$. Use this relation to get the answer.

Complete step by step answer:

- In an atom, electrons and nucleus are present along with other subatomic particles. Electrons revolve around the nucleus because of the nuclear force of attraction.- For an electron to undergo electronic transition from lower energy level to higher energy level, energy is required. If an electron jumps from higher energy level to lower energy level then, energy is given out in the form of photons.- In the question, we can see that all transitions are taking place from higher energy levels to lower energy levels. Hence, photons are emitted out in each of the transitions.- For calculating wavelength of a photon, we have the formula, $\dfrac{1}{\lambda }=R\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)$- For minimum wavelength, $\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)$ should be maximum.(A) n=6 to n=4\[\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)=\dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{6}^{2}}}=\dfrac{1}{16}-\dfrac{1}{36}=0.035\](B) n=4 to n=2\[\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)=\dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}}=\dfrac{1}{4}-\dfrac{1}{16}=0.1875\](C) n=3 to n=1\[\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)=\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}}=\dfrac{1}{1}-\dfrac{1}{9}=0.8889\](D) n=2 to n=1\[\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)=\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}}=\dfrac{1}{1}-\dfrac{1}{4}=0.75\]- The transition n=3 to n=1 has the highest value of $\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)$ that is 0.8889.- Therefore, n = 3 to n = 1 transition will have the least wavelength.

So, the correct answer is “Option C”.

Note: Remember energy is inversely proportional to wavelength. For an electron to undergo electronic transition from lower energy level to higher energy level, energy is absorbed. If an electron jumps from higher energy level to lower energy level then, energy is released in the form of photons. So, if in a transition, energy released or absorbed is high then the wavelength is minimum.