An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom?
An atom having atomic mass number 13 and number of neutrons 7 .
i.e., $$A=13, n=7$$
As we know that,
$$\begin{array}{ll} \text { As we know that, } & A=n+p \\ \therefore & p=A-n=13-7=6 \end{array}$$
Hence, $$Z=p=6$$
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.