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Structure of Atom question

2017 · 8 Apr · Shift 1 · Q22
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Structure of Atom question

2017 · 8 Apr · Shift 1 · Q22

JEE MainChemistryStructure of AtomMCQ+4 / −1
If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He+He^+He+ is :
  1. A
    5A9{{5A} \over 9}95A​
  2. B
    9A5{{9A} \over 5}59A​
  3. C
    36A5{{36A} \over 5}536A​
  4. D
    36A7{{36A} \over 7}736A​
View written solutionFree

Correct answer: D

  1. Use the Rydberg formula

For a hydrogen-like species,

1λ=RZ2(1n12−1n22)\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)λ1​=RZ2(n12​1​−n22​1​)

where ZZZ is atomic number.

  1. Find the shortest wavelength in Lyman series of hydrogen

For hydrogen, Z=1Z=1Z=1.

Lyman series means transitions to n1=1n_1=1n1​=1. The shortest wavelength corresponds to the largest energy gap, i.e. n2→∞n_2 \to \inftyn2​→∞.

So,

1A=R(112−1∞2)=R\frac{1}{A} = R\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right)=RA1​=R(121​−∞21​)=R

Hence,

A=1RA = \frac{1}{R}A=R1​
  1. Find the longest wavelength in Paschen series of He+He^+He+

For He+He^+He+, Z=2Z=2Z=2.

Paschen series means transitions to n1=3n_1=3n1​=3. The longest wavelength corresponds to the smallest energy gap, i.e. first line of the series: n2=4→n1=3n_2=4 \to n_1=3n2​=4→n1​=3.

Thus,

1λ=R(2)2(132−142)=4R(19−116)\frac{1}{\lambda} = R(2)^2\left(\frac{1}{3^2} - \frac{1}{4^2}\right) =4R\left(\frac{1}{9}-\frac{1}{16}\right)λ1​=R(2)2(321​−421​)=4R(91​−161​)

Now,

19−116=16−9144=7144\frac{1}{9}-\frac{1}{16} = \frac{16-9}{144} = \frac{7}{144}91​−161​=14416−9​=1447​

Therefore,

1λ=4R⋅7144=28R144=7R36\frac{1}{\lambda} = 4R \cdot \frac{7}{144} = \frac{28R}{144} = \frac{7R}{36}λ1​=4R⋅1447​=14428R​=367R​

So,

λ=367R\lambda = \frac{36}{7R}λ=7R36​

Since A=1RA=\frac{1}{R}A=R1​,

λ=36A7\lambda = \frac{36A}{7}λ=736A​
  1. Match with options
λ=36A7\boxed{\lambda = \frac{36A}{7}}λ=736A​​

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.

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