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Structure of Atom question

2004 · Q100
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Structure of Atom question

2004 · Q100

NEETChemistryStructure of AtomMCQ+4 / −1
The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in hydrogen atom will be (Given ionization energy of H = 2.18 ×\times× 10−-−18 J atom−-−1 and h = 6.625 ×\times× 10−-−34 J s)
  1. A
    1.54 ×\times× 1015 s−-−1
  2. B
    1.03 ×\times× 1015 s−-−1
  3. C
    3.08 ×\times× 1015 s−-−1
  4. D
    2.00 ×\times× 1015 s−-−1
View written solutionFree

Correct answer: C

E = hvvv or vvv = E/h

For H atom, E = −21.76×10−19n2J atm−1{{ - 21.76 \times {{10}^{ - 19}}} \over {{n^2}}}J\,at{m^{ - 1}}n2−21.76×10−19​Jatm−1

ΔE=−21.76×10−19(142−112)\Delta E = - 21.76 \times {10^{ - 19}}\left( {{1 \over {{4^2}}} - {1 \over {{1^2}}}} \right)ΔE=−21.76×10−19(421​−121​)

= 20.40 ×\times× 10-19 J atm-1

vvv = 20.40×10−196.626×10−34{{20.40 \times {{10}^{ - 19}}} \over {6.626 \times {{10}^{ - 34}}}}6.626×10−3420.40×10−19​ = 3.079 ×\times× 1015 s-1

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