Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Structure of Atom question

2009 · Shift 0 · Q25
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Structure of Atom
  5. /2009 · Shift 0 · Q25

Structure of Atom question

2009 · Shift 0 · Q25

JEE MainChemistryStructure of AtomMCQ+4 / −1
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 x 103 ms−1 (Mass of proton = 1.67 ×\times× 10-27 kg and h = 6.63 ×\times× 10-34 Js) :
  1. A
    0.40 nm
  2. B
    2.5 nm
  3. C
    14.0 nm
  4. D
    0.32 nm
View written solutionFree

Correct answer: A

  1. Use de Broglie wavelength relation

    The wavelength associated with a moving particle is λ=hmv\lambda = \frac{h}{mv}λ=mvh​

  2. Substitute the given values

    For the proton: h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s m=1.67×10−27 kgm = 1.67 \times 10^{-27}\,\text{kg}m=1.67×10−27kg v=1.0×103 m s−1v = 1.0 \times 10^3\,\text{m s}^{-1}v=1.0×103m s−1

    So, λ=6.63×10−34(1.67×10−27)(1.0×103)\lambda = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27})(1.0 \times 10^3)}λ=(1.67×10−27)(1.0×103)6.63×10−34​

  3. Simplify the denominator

    1.67×10−27×1.0×103=1.67×10−241.67 \times 10^{-27} \times 1.0 \times 10^3 = 1.67 \times 10^{-24}1.67×10−27×1.0×103=1.67×10−24

    Therefore, λ=6.63×10−341.67×10−24\lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-24}}λ=1.67×10−246.63×10−34​

  4. Calculate the value

    λ=(6.631.67)×10−10\lambda = \left(\frac{6.63}{1.67}\right) \times 10^{-10}λ=(1.676.63​)×10−10 λ≈3.97×10−10 m\lambda \approx 3.97 \times 10^{-10}\,\text{m}λ≈3.97×10−10m

  5. Convert meter to nanometer

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

    λ=3.97×10−10 m=0.397 nm\lambda = 3.97 \times 10^{-10}\,\text{m} = 0.397\,\text{nm}λ=3.97×10−10m=0.397nm

    λ≈0.40 nm\lambda \approx 0.40\,\text{nm}λ≈0.40nm

  6. Check options

    • A: 0.40 nm0.40\,\text{nm}0.40nm ✅
    • B: 2.5 nm2.5\,\text{nm}2.5nm ❌
    • C: 14.0 nm14.0\,\text{nm}14.0nm ❌
    • D: 0.32 nm0.32\,\text{nm}0.32nm ❌

Therefore, the correct option is A.

PreviousNext

More from Structure of Atom

  • The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol−1. The energy required to excite the electron in the atom from n = 1 to n = 2 is2008 · MCQ
  • Which one of the following constitutes a group of the isoelectronic species?2008 · MCQ
  • Which of the following sets of quantum numbers represents the highest energy of an atom?2007 · MCQ
  • Uncertainty in the position of an electron (mass = 9.1 × 10-31 kg) moving with a velocity 300 ms-1, accurate upto 0.001% will be (h = 6.63 × 10-34 Js)2006 · MCQ
  • According to Bohr's theory, the angular momentum of an electron in 5th orbit is2006 · MCQ
  • Which of the following sets of ions represents a collection of isoelectronic species?2006 · MCQ
  • Pick out the isoelectronic structure from the following : ​(i)CH3​+(ii)H3​O+(iii)NH3​(iv)CH3​−​…2005 · MCQ
  • Of the following sets which one does NOT contain isoelectronic species?2005 · MCQ