According to de-Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of $100 \mathrm{~km} / \mathrm{h}$. Calculate the wavelength of the ball and explain why it does not show wave nature.
Given,
$$\begin{aligned} m & =100 \mathrm{~g}=0.1 \mathrm{~kg} \\ v & =100 \mathrm{~km} / \mathrm{h}=\frac{100 \times 1000}{60 \times 60}=\frac{1000}{36} \mathrm{~ms}^{-1} \end{aligned}$$
From de-Broglie equation, wavelength, $\lambda=\frac{h}{m v}$
$$\lambda=\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}}{0.1 \mathrm{~kg} \times \frac{1000}{36} \mathrm{~ms}^{-1}}=238.5 \times 10^{-36} \mathrm{~m}$$
As the wavelength is very small so wave nature cannot be detected.
What is the experimental evidence in support of the idea that electronic energies in an atom are quantized?
The line spectrum of any element has lines corresponding to definite wavelengths. Lines are obtained as a result of electronic transitions between the energy levels. Hence, the electrons in these levels have fixed energy, i.e., quantized values.
Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.
From de-Broglie equation, wavelength, $\lambda=\frac{h}{m v}$
For same wavelength for two different particles, i.e., electron and proton, $m_1 v_1=m_2 v_2$ ( $h$ is constant). Lesser the mass of the particle, greater will be the velocity. Hence, electron will have higher velocity.
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.