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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2020 · 8 January · Morning Shift · Chemistry · Structure of Atom

For the Balmer series, in the spectrum of H atom , - v = R _ H \ 1...

For the Balmer series, in the spectrum of HH atom, v−=RH{1n12−1n22},\overset{-}{v} = {R}_{H}\left\{1{n}_{1}^{2} - 1{n}_{2}^{2}\right\}, the correct statements among \left(I\right) to \left(IV\right)are,<br/>(I)are,<br />\left(I\right) As wavelength decreases, the lines in the series converge.
\left(II\right)TheintegerThe integer{n}_{1}isequaltois equal to2/<br/>(III)/<br />\left(III\right) The lines of the longest wavelength correspond to n2=3{n}_{2} = 3.
\left(IV\right)$ The ionization energy of hydrogen can be calculated from the wave number of these lines.

Options

  1. A(I),(III),(IV)\left(I\right), \left(III\right), \left(IV\right)
  2. B(I),(II),(III)\left(I\right), \left(II\right), \left(III\right)
  3. C(I),(II),(IV)\left(I\right), \left(II\right), \left(IV\right)
  4. D(II),(III),(IV)\left(II\right), \left(III\right), \left(IV\right)

Correct answer

B. (I),(II),(III)\left(I\right), \left(II\right), \left(III\right)

Step-by-step solution

\left(I\right), \left(II\right)and \left(III\right) option is correct.

ν−=RH[1n12−1n22]\overset{-}{ν} = {R}_{H}\left[\dfrac{1}{{{n}_{1}}^{2}} - \dfrac{1}{{{n}_{2}}^{2}}\right]

For balmer series, n1=2{n}_{1} = 2

For the longest wavelength, V¯\overset{¯}{V},

n2=3{n}_{2} = 3.
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