Electrons with paired spins, one electron has $↑$ direction and the other has $↓$ direction, these are denoted by:

m_{s} = + \frac{1}{2}

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m_{s} = \frac{1}{2}

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m_{s} = + \frac{1}{2}, m_{s} = - \frac{1}{2}

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m_{s} = - \frac{1}{2}

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Solution

The correct option is A $m_{s} = + \frac{1}{2}, m_{s} = - \frac{1}{2}$
The spin quantum number $(m_{s}$ denotes the direction of spin of electron.)
Therefore, the $↑$ and the $↓$ direction is denoted as:
$m_{s}$ = + $\frac{1}{2}$ and $m_{s} = - \frac{1}{2}$ respectively.

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Question 18

Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals.

(a) (i) $n = 3, l = 2, m_{l} = - 2, m_{s} = - \frac{1}{2}$ (ii) $n = 3, l = 2, m_{l} = - 1, m_{s} = - \frac{1}{2}$

(b) (i) $n = 3, l = 1, m_{l} = 1, m_{s} = + \frac{1}{2}$ (ii) $n = 3, l = 2, m_{l} = - 1, m_{s} = + \frac{1}{2}$

(d) (i) $n = 3, l = 2, m_{l} = + 2, m_{s} = - \frac{1}{2}$ (ii) $n = 3, l = 2, m_{l} = + 2, m_{s} = + \frac{1}{2}$

A body has an initial velocity of 4 $m s^{- 1}$ and deceleration of 4 $m s^{- 2}$ , then what is its final velocity after traveling 2m?

A body has an initial velocity 5 $m s^{- 1}$ . It is accelerating at 2 $m s^{- 2}$ . It's velocity after 10 second is

n = 4, l = 2, m_{1} = - 2, m_{s} = - 1 / 2

n = 3, l = 2, m_{1} = 1, m_{s} = + 1 / 2

n = 4, l = 1, m_{1} = 0, m_{s} = + 1 / 2

n = 3, l = 2, m_{1} = - 2, m_{s} = - 1 / 2

n = 3, l = 1, m_{1} = - 1, m_{s} = + 1 / 2

n = 4, l = 1, m_{1} = 0, m_{s} = + 1 / 2

20 m s^{- 1}

20 m s^{- 1}

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