A hypothetical electromagnetic wave is shown in figure. Find out the wavelength of the radiation.
Wavelength is the distance between two successive peaks or two successive throughs of a wave.
Therefore,
$$\begin{aligned} \lambda=4 \times 2.16 \mathrm{pm} & =8.64 \mathrm{~pm} \\ & =8.64 \times 10^{-12} \mathrm{~m} \quad\left[\because 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right] \end{aligned}$$
Chlorophyll present in green leaves of plants absorbs light at $4.620 \times 10^{14} \mathrm{~Hz}$. Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to?
$$\begin{aligned} &\text { Wavelength, }\quad \lambda=\frac{C}{V} \end{aligned}$$
$$\begin{aligned} \text { Given, } \quad \text { frequency } \nu & =4.620 \times 10^{14} \mathrm{~Hz} \text { or } 4.620 \times 10^{14} \mathrm{~s}^{-1} \\ \lambda & =\frac{c}{v}=\frac{3 \times 10^8 \mathrm{~ms}^{-1}}{4.620 \times 10^{14} \mathrm{~s}^{-1}} \\ & =0.6494 \times 10^{-6} \mathrm{~m} \\ & =649.4 \mathrm{~nm}\quad \left[\because 1 \mathrm{~nm}=10^{-6} \mathrm{~m}\right] \end{aligned}$$
Thus, it belongs to visible region.
What is the difference between the terms orbit and orbital?
Difference between the terms orbit and orbital are as below
| Orbit | Orbital |
|---|---|
| An orbit is a well defined circular path around the nucleus in which the electrons revolve. | An orbital is the three dimensional space around the nucleus within which the probability of finding an electron is maximum (upto 90%). |
| All orbits are circular and disc like. |
Different orbitals have different shapes. |
| The concept of an orbit is not in accordance with the wave character of electrons with uncertainty principle. | The concept of an orbital is in accordance with the wave character of electrons and uncertainty principle. |
| The maximum number of electrons in any orbit is given by $$2 n^2$$ where $$n$$ is the number of the orbit. | The maximum number of electrons present in any orbital is two. |
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.