Table-tennis ball has mass 10 g and speed of $90 \mathrm{~m} / \mathrm{s}$. If speed can be measured within an accuracy of $4 \%$, what will be the uncertainty in speed and position?
$$\begin{aligned} & \text { Given that, speed }=90 \mathrm{~m} / \mathrm{s} \\ & \qquad \text { mass }=10 \mathrm{~g}=10 \times 10^{-3} \mathrm{~kg} \\ & \text { Uncertainty in speed }(\Delta v)=4 \% \text { of } 90 \mathrm{~ms}^{-1}=\frac{4 \times 90}{100}=3.6 \mathrm{~ms}^{-1} \end{aligned}$$
From Heisenberg uncertainty principle,
$$\begin{aligned} \Delta x \cdot \Delta v & =\frac{h}{4 \pi m} \\ \text{or}\quad \Delta x & =\frac{h}{4 \pi m \Delta v} \end{aligned}$$
Uncertainty in position,
$$\begin{aligned} \Delta x & =\frac{6.626 \times 10^{-34} \mathrm{kgm}^2 \mathrm{~s}^{-1}}{4 \times 3.14 \times 10 \times 10^{-3} \mathrm{~kg} \times 3.6 \mathrm{~ms}^{-1}} \\ & =1.46 \times 10^{-33} \mathrm{~m} \end{aligned}$$
The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example.
$$\begin{aligned} &\text { If uncertainty principle is applied to an object of mass, say about a milligram }\left(10^{-6} \mathrm{~kg}\right) \text {, then }\\ &\begin{aligned} \Delta \cdot \Delta x & =\frac{h}{4 \pi m} \\ \Delta v \cdot \Delta x & =\frac{6.626 \times 10^{-34} \mathrm{kgm}^2 \mathrm{~s}^{-1}}{4 \times 3.14 \times 10^{-6} \mathrm{~kg}} \\ & =0.52 \times 10^{-28} \mathrm{~m}^2 \mathrm{~s}^{-1} \end{aligned} \end{aligned}$$
The value of $\Delta v \cdot \Delta x$ obtained is extremely small and is insignificant. Therefore, for milligram-sized or heavier objects, the associated uncertainties are hardly of any real consequence.
Hydrogen atom has only one electron. So, mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?
In hydrogen atom, the energy of an electron is determined by the value of $n$ and in multielectron atom, it is determined by $n+l$. Hence, for a given principal quantum, number electrons of $s, p, d$ and $f$-orbitals have different energy (for $s, p, d$ and $f, l=0,1,2$ and 3 respectively).
Match the following species with their corresponding ground state electric configuration.
| Atom / Ion | Electronic configuration | ||
|---|---|---|---|
| A. | Cu | 1. | $$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} $$ |
| B. | Cu$$^{2+}$$ | 2. | $$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4s^2 $$ |
| C. | Zn$$^{2+}$$ | 3. | $$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4s^1 $$ |
| D. | Cr$$^{3+}$$ | 4. | $$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{9} $$ |
| 5. | $$ 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{3} $$ |
A. $\rightarrow(3)$
B. $\rightarrow(4)$
C. $\rightarrow(1)$
D. $\rightarrow(5)$
A. $\mathrm{Cu}(\mathrm{Z}=29) \quad: \quad 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1$
B. $\mathrm{Cu}^{2+}(Z=29) \quad: \quad 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^9$
C. $Z n^{2+}(Z=30) \quad: \quad 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10}$
D. $\mathrm{Cr}^{3+}(\mathrm{Z}=24) \quad: \quad 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^3$
Match the quantum numbers with the information provided by these.
| Quantum number | Information provided | ||
|---|---|---|---|
| A. | Principal quantum number | 1. | Orientation of the orbital |
| B. | Azimuthal quantum number | 2. | Energy and size of orbital |
| C. | Magnetic quantum number | 3. | Spin of electron |
| D. | Spin quantum number | 4. | Shape of the orbital |
A. $\rightarrow$ (2)
B. $\rightarrow(4)$
C. $\rightarrow(1)$
D. $\rightarrow$ (3)
A. Principal quantum number is the most important quantum number as it determines the size and to large extent the energy of the orbital.
B. Azimuthal quantum number determines the angular momentum of the electron and defines the three-dimensional shape of the orbital.
C. Magnetic quantum number gives information about the spatial orientation of orbitals with respect to a standard set of coordinate axes.
D. Spin quantum number arises from the spectral evidence that an electron in its motion around the nucleus in an orbit also rotates or spin about its own axis.