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JEE Main 201912 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2019 · 12 January · Evening Shift · Chemistry · Structure of Atom

If the de Broglie wavelength of the electron in n ^ th Bohr orbit...

If the de Broglie wavelength of the electron in nth{n}^{th} Bohr orbit in a hydrogenic atom is equal to 1.5πa01.5\pi{a}_{0} (a0( {a}_{0} is Bohr radius)), then the value of nz\dfrac{n}{z} is:

Options

  1. A1.501.50
  2. B1.01.0
  3. C0.40.4
  4. D0.75

Correct answer

D. 0.75

Step-by-step solution

2\pir=nλ2\pir = n\lambda

2πa0n2Z=nλ2\pi{a}_{0}\dfrac{{n}^{2}}{Z} = n\lambda

2πa0n2Z=n×1.5πa02\pi{a}_{0}\dfrac{{n}^{2}}{Z} = n \times 1.5\pi{a}_{0}

nZ=1.52=0.75\dfrac{n}{Z} = \dfrac{1.5}{2} = 0.75
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