Atoms & Nuclei

JEE Main level, numerical answer, medium

109 of 239

Question 109 of 239

A particular hydrogen - like ion emits the radiation of frequency 3×1015Hz3\times {10}^{15}Hz when it makes transition from n=2n=2 to n=1.n=1. The frequency of radiation emitted in transition from n=3n=3 to n=1n=1 is x9×1015Hz,\dfrac{x}{9}\times {10}^{15}Hz, when x=x= _____.

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Answer: 32

For Hydrogen like atom, energy emitted is E=−13.6Z2(1ni2−1nf2)E=-13.6{Z}^{2}\left(\dfrac{1}{{n}_{i}^{2}}-\dfrac{1}{{n}_{f}^{2}}\right)
⇒E=C(1nf2−1ni2)\Rightarrow E=C\left(\dfrac{1}{{n}_{f}^{2}}-\dfrac{1}{{n}_{i}^{2}}\right)
⇒hν=C[1nf2−1ni2]\Rightarrow h\nu =C\left[\dfrac{1}{{n}_{f}^{2}}-\dfrac{1}{{n}_{i}^{2}}\right]
Therefore, the ratiov1v2=[1nf2−1ni2]2−1[1nf2−1ni2]3−1\dfrac{{v}_{1}}{{v}_{2}}=\dfrac{{\left[\dfrac{1}{{n}_{f}^{2}}-\dfrac{1}{{n}_{i}^{2}}\right]}_{2-1}}{{\left[\dfrac{1}{{n}_{f}^{2}}-\dfrac{1}{{n}_{i}^{2}}\right]}_{3-1}}
=[11−14][11−19]=3489=\dfrac{\left[\dfrac{1}{1}-\dfrac{1}{4}\right]}{\left[\dfrac{1}{1}-\dfrac{1}{9}\right]}=\dfrac{\dfrac{3}{4}}{\dfrac{8}{9}}
=34×98=\dfrac{3}{4}\times \dfrac{9}{8}
⇒v1v2=2732\Rightarrow \dfrac{{v}_{1}}{{v}_{2}}=\dfrac{27}{32}
⇒v2=3227v1=3227×3×1015Hz=329×1015Hz\Rightarrow {v}_{2}=\dfrac{32}{27}{v}_{1}=\dfrac{32}{27}\times 3\times {10}^{15}Hz=\dfrac{32}{9}\times {10}^{15}Hz
Hence, x=32x=32.

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