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Structure of Atom question

2004 · Shift 0 · Q57
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Structure of Atom question

2004 · Shift 0 · Q57

JEE MainChemistryStructure of AtomMCQ+4 / −1
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 ×\times× 107 m-1)
  1. A
    406 nm
  2. B
    192 nm
  3. C
    91 nm
  4. D
    9.1 ×\times× 10-8 nm
View written solutionFree

Correct answer: C

  1. Use the Rydberg formula for hydrogen:
1λ=R(1n12−1n22)\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)λ1​=R(n12​1​−n22​1​)

where:

  • R=1.097×107 m−1R = 1.097 \times 10^7\ \text{m}^{-1}R=1.097×107 m−1
  • final state n1=1n_1 = 1n1​=1
  • initial state is infinity, so n2=∞n_2 = \inftyn2​=∞
  1. Substitute the values:
1λ=1.097×107(1−1∞2)\frac{1}{\lambda} = 1.097 \times 10^7 \left(1 - \frac{1}{\infty^2}\right)λ1​=1.097×107(1−∞21​)

Since

1∞2=0\frac{1}{\infty^2} = 0∞21​=0

we get

1λ=1.097×107\frac{1}{\lambda} = 1.097 \times 10^7λ1​=1.097×107
  1. Calculate the wavelength:
λ=11.097×107 m\lambda = \frac{1}{1.097 \times 10^7}\ \text{m}λ=1.097×1071​ m λ≈9.115×10−8 m\lambda \approx 9.115 \times 10^{-8}\ \text{m}λ≈9.115×10−8 m
  1. Convert into nm:

Since 1 nm=10−9 m1\ \text{nm} = 10^{-9}\ \text{m}1 nm=10−9 m,

λ=9.115×10−8×109 nm\lambda = 9.115 \times 10^{-8} \times 10^9\ \text{nm}λ=9.115×10−8×109 nm λ≈91.15 nm\lambda \approx 91.15\ \text{nm}λ≈91.15 nm
  1. Match with the options:
  • A: 406 nm406\ \text{nm}406 nm
  • B: 192 nm192\ \text{nm}192 nm
  • C: 91 nm91\ \text{nm}91 nm
  • D: 9.1×10−8 nm9.1 \times 10^{-8}\ \text{nm}9.1×10−8 nm

So the correct option is:

C: 91 nm\boxed{\text{C: } 91\ \text{nm}}C: 91 nm​
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