Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron.
Neutron being neutral will not show deflection from the path on passing through an electric field.
Proton, cathode rays and electron being the charged particle will show deflection from the path on passing through an electric field.
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,
$$\begin{aligned} \text{Wave number,}\quad & \bar{v}=109677\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right] \mathrm{cm}^{-1} \\ \text{Given,} \quad & n_i=2 \text { and } n_f=4 \quad \text{(Transition in Balmer series)}\\ & \bar{v}=109677\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \mathrm{cm}^{-1} \\ \Rightarrow \quad & \bar{v}=109677\left[\frac{1}{4}-\frac{1}{16}\right] \mathrm{cm}^{-1} \\ \Rightarrow \quad & \bar{v}=109677 \times\left[\frac{4-1}{16}\right] \mathrm{cm}^{-1} \\ \Rightarrow \quad & \bar{v}=20564.44 \mathrm{~cm}^{-1} \end{aligned}$$