Atoms & Nuclei

JEE Main level, numerical answer, medium

116 of 239

Question 116 of 239

As per given figure A,BA,B and CC are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengths(i.e.λ1λ2)\left(i.e.\dfrac{{\lambda }_{1}}{{\lambda }_{2}}\right) is 74n\dfrac{7}{4n}, then the value of nn will be

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Answer: 5

For λ1{\lambda }_{1}
1λ1=RZ2(122−132)=R(536)...(i)\dfrac{1}{{\lambda }_{1}}=R{Z}^{2}(\dfrac{1}{{2}^{2}}-\dfrac{1}{{3}^{2}})=R\left(\dfrac{5}{36}\right)...(i)
For λ2{\lambda }_{2},
1λ2=RZ2(132−142)=R(7144)...(ii)\dfrac{1}{{\lambda }_{2}}=R{Z}^{2}(\dfrac{1}{{3}^{2}}-\dfrac{1}{{4}^{2}})=R\left(\dfrac{7}{144}\right)...(ii)
Dividing (ii) by (i)
λ1λ2=7144536=720\dfrac{{\lambda }_{1}}{{\lambda }_{2}}=\dfrac{\dfrac{7}{144}}{\dfrac{5}{36}}=\dfrac{7}{20}
So,
74n=720⇒n=5\dfrac{7}{4n}=\dfrac{7}{20} \Rightarrow n=5

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