Wavelengths of different radiations are given below.
$\lambda(\mathrm{A})=300 \mathrm{~nm} \lambda$
$\lambda (\mathrm{B}) =300 \mu \mathrm{m} \lambda$
$\lambda (\mathrm{C})=3 \mathrm{~nm} \lambda$
$\lambda (\mathrm{D})=30 \mathop A\limits^o$
Arrange these radiations in the increasing order of their energies.
$$\begin{array}{ll} \lambda(A)=300 \mathrm{~nm}=300 \times 10^{-9} \mathrm{~m}, & \lambda(\mathrm{B})=300 \mu \mathrm{m}=300 \times 10^{-6} \mathrm{~m} \\ \lambda(\mathrm{C})=3 \mathrm{~nm}=3 \times 10^{-9} \mathrm{~m}, & \lambda(\mathrm{D})=30 \mathop A\limits^o=30 \times 10^{-10} \mathrm{~m}=3 \times 10^{-9} \mathrm{~m} \end{array}$$
$\begin{array}{ll}\text { Energy, } & E=\frac{h c}{\lambda} \\ \text { Therefore, } & E \propto \frac{1}{\lambda}\end{array}$
Increasing order of energy is $B< A< C=D$
The electronic configuration of valence shell of Cu is $3 d^{10} 4 s^1$ and not $3 d^9 4 s^2$. How is this configuration explained?
Configurations either exactly half-filled or fully filled orbitals are more stable due to symmetrical distribution of electrons and maximum exchange energy. In $3 d^{10} 4 s^1, d$-orbitals are completely filled and s-orbital is half-filled. Hence, it is more stable configuration.
The Balmer series in the hydrogen spectrum corresponds to the transition from $n_1=2$ to $n_2=3,4, \ldots$. . This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to $n=4$ orbit. $$(R_{\mathrm{H}}=109677 \mathrm{~cm}^{-1})$$
From Rydberg formula,
$$\begin{aligned} \text{Wave number,}\quad & \bar{v}=109677\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right] \mathrm{cm}^{-1} \\ \text{Given,} \quad & n_i=2 \text { and } n_f=4 \quad \text{(Transition in Balmer series)}\\ & \bar{v}=109677\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \mathrm{cm}^{-1} \\ \Rightarrow \quad & \bar{v}=109677\left[\frac{1}{4}-\frac{1}{16}\right] \mathrm{cm}^{-1} \\ \Rightarrow \quad & \bar{v}=109677 \times\left[\frac{4-1}{16}\right] \mathrm{cm}^{-1} \\ \Rightarrow \quad & \bar{v}=20564.44 \mathrm{~cm}^{-1} \end{aligned}$$
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.