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BITSAT 2017
Exam
BITSAT
Year
2017
Subject
Chemistry
Chapter
Structure of Atom

BITSAT 2017 · Chemistry · Structure of Atom

The orbital angular momentum of an electron is 2s & orbital is

The orbital angular momentum of an electron is 2s2s& orbital is

Options

  1. Ah4π\dfrac{h}{4\pi}
  2. Bzero
  3. Ch2π\dfrac{h}{2\pi}
  4. D2⋅h2π\sqrt{2} \cdot \dfrac{h}{2\pi}

Correct answer

B. zero

Step-by-step solution

For ss-subshell l=0l = 0

Orbital angular momentum (mvr)=h2πl(l+1)( mvr ) = \dfrac{h}{2\pi}\sqrt{l ( l + 1 )}

For 2s2s orbital, ll (azimuthal quantum number) =0= 0
∴∴ orbital angular momentum =h2π0(0+1)= \dfrac{h}{2\pi}\sqrt{0 ( 0 + 1 )}

=h2π0=0= \dfrac{h}{2\pi}\sqrt{0} = 0

Thus, the orbital angular momentum of an electron in 2s2s -orbital is& Zero.

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