From Rydberg's relation:

Concept attached
$$r_n=Cn^2$$
$$ r_1 = C $$
$$r_n= r_1 n^2$$
$$A_n= \pi r_n ^2 = \pi r_1 ^2 n^4$$
$$ log(\cfrac{A_n}{A_1}) = 4 logn +K$$
Slope =4, Linear graph.

$$\Delta E_{h}=\dfrac{3}{4}\times 13.6\ eV =$$ Energy released by H atom. 

$${ n }^{ th }$$ shell can hold upto $$2{ n }^{ 2 }$$ electrons
Total number of possible elements would be $$\sum _{ 1 }^{ n }{ 2{ n }^{ 2 } } $$$$=\dfrac { n }{ 3 } \left( n+1 \right) \left( 2n+1 \right) $$

Total no. of wavelengths emitted $$=\ ^n C _2= \cfrac{n(n-1)}{2}$$
$$^n C _2 = 10$$
$$ n^2 -n -20 =0$$
$$ n=-4 \ or \:5$$
$$n=5$$

Angular momentum, $$L=mvr_n=\dfrac{nh}{2\pi}\propto n$$
Hence, (ii) is correct.
Velocity of $$2^{nd}$$ orbit is less so kinetic energy is also less. 

We know that: 

$$ mvR=nh$$
$$V_o = \sqrt{\cfrac{GM}{R}}$$
Substitute:

Potential energy at a distance $$R$$ is given as, $$U=\dfrac{ke^2}{2R^3}$$

Beta and gamma ray can be detected by GM counter but alpha partciles cannot be detected as they cannot pass through the Aluminium window.

In X-ray spectroscopy, K-alpha emission lines result when an electron transitions to the innermost "K" shell (principal quantum number 1) from a 2p orbital of the second or "L" shell (with principal quantum number 2). 
 $$ \cfrac{1}{\lambda}= Z^2 R(1 - \cfrac{1}{2^2}) = \cfrac{3RZ^2}{4}$$
$$ \lambda _A =  \cfrac{4}{1875R} = \cfrac{4}{3 R Z^2}$$
$$ Z_A =25$$

Similarly,
$$ \lambda _B =  \cfrac{1}{675R} = \cfrac{4}{3 R Z_B ^2}$$
$$Z_B = 30$$

Total no. of levels $$= n-1$$
$$ n \rightarrow 2$$
Put $$n' = n-1 $$ in the formula attached
$$ \cfrac{(n-1) (n-2)}{2} = 10$$
$$ (n-1)(n-2) = 20$$
$$\Rightarrow  n=6$$

As per the data shown in the figure, and the wavelenghts that can be determined using the Rydberg formula, the correct ans. can be obtained
$$\cfrac{1}{\lambda} = Z^2 R (\cfrac{1}{n_1 ^2}- \cfrac{1}{n_2 ^2})$$

Radius
$$r = cn^2$$

Angular Momentum Quantisation:
$$ mvr = \cfrac{nh}{2 \pi}$$                    ------------(1)

Angular Momentum Quantisation:
$$ mvr = \cfrac{nh}{2 \pi}$$                    ------------(1)

For hydrogen atom , radius $$r_n \propto n^2$$

 A line emission spectrum is obtained when the emission occurs from an atom, where the lines are usually unique and distinct in nature, while a band emission spectrum is in the form of continuous bands that are obtained from molecules.
In simple terms, a line emission spectrum consists a variety of frequencies of electromagnetic radiations, generally emitted by elements, atoms or compound molecules at their lower energy states. Each emission spectrum in a line emission is unique and distinct in nature.
A band emission spectrum is generally obtained from various line spectra. These are closely spaced together in a continuous state. Unlike line spectrum, the band spectra are obtained from various line spectra, while they differ in characteristics and properties. In physics, band emission is described as a fraction of the total emission from a blackbody, which comes in different intervals among certain wavelengths 

One means by which a continuous spectrum can be produced is by thermal emission from a black body.
Bohr's Quantisation principle clearly states that the energy levels in the atom are quantised and hence only certain wavelenghts can be emitted and absorbed by the atom.

Radius of n th orbit : $$R \propto \dfrac{n^2}{Z}= n^2   ...(1)$$ (for hydrogen atom $$Z=1$$)

For an atom following Bohr's model, the radius is given by $$\displaystyle r_m=\frac{r_0m^2}{Z}$$ 

Given, electron in the positronium has reduced mass $$\dfrac { { m }_{ e } }{ 2 }$$
And orbital radius of ground state of hydrogen atom is $${ a }_{ 0 }$$.
Therefore, for positronium, $${ r }_{ n }=2{ a }_{ 0 }{ n }^{ 2 }$$
Hence, orbital radius of first excited level of positronium  is $$8{ a }_{ 0 }$$.

Laws of Motion:
Newtons second law: $$F = ma$$
Without any force, the atom cannot accelerate.
Conservation of Momentum:
The linear momentum of the atom can't change without any external force acting on it.

Radius $$\propto n^2$$
Hence, $$R_1:R_2:R_3 = 1:4:9$$

Given, $$47.2\ eV$$ is required to excite an electron from $$n=2$$ to $$n=3$$
$$\Rightarrow  \dfrac { 13.6 { Z }^{ 2 } }{ 4 } -\dfrac { 13.6 { Z }^{ 2 } }{ 9 } =\dfrac { 5\times 13.6 { Z }^{ 2 } }{ 36 } =47.2$$ 

Absorption occurs when electron excites to higher level.
For Balmer, electron has to excite from $$n=2$$.

Minimum radius will be when $$n=1$$:
$$mvR_\circ{}=2 \dfrac{h}{2\pi}\Rightarrow vR_\circ{}= \dfrac{h}{m\pi}$$...(i)

$$\dfrac {mv^{2}}{r} = \dfrac {KQ^{2}}{r^{2}}$$
and
$$mvr = \dfrac {nh}{2\pi}\implies mv = \dfrac{nh}{2 \pi r}$$

Radius of ground state, $$r = \dfrac{h^2}{4\pi ^2 mKe^2}  $$

Centripetal force$$=\dfrac { m{ v }^{ 2 } }{ { a }_{ \circ  } } \rightarrow (1)$$

Let the velocity of the electron be $$v$$.
Angular momentum of electron  $$mvr = \hbar$$   (given)
Net electrostatic force acting on the electron  $$F_e = \dfrac{e^2}{4\pi\epsilon_o r^2} - \dfrac{e^2}{4\pi\epsilon_o (2r^2)} = \dfrac{3}{4}\dfrac{e^2}{4\pi\epsilon_o r^2}$$
Centrifugal force  $$F_c = \dfrac{mv^2}{r} = \dfrac{(mvr)^2}{r\times mr^2} = \dfrac{\hbar^2}{mr^3}$$
But   $$F_c = F_e$$
$$\therefore$$  $$\dfrac{\hbar^2}{mr^3} = \dfrac{3}{4}\dfrac{e^2}{4\pi\epsilon_o r^2}$$
$$\implies \ r = \dfrac{4}{3}\dfrac{4\pi\epsilon_o \hbar^2}{me^2} = \dfrac{4}{3}a_B$$

From Bohr's theory
$$mvr=\displaystyle\frac{nh}{2\pi}$$
For first orbit $$n=1$$
$$m(r\omega )r=\displaystyle\frac{1\times h}{2\pi}$$
$$\Rightarrow mr^22\pi f=\displaystyle\frac{h}{2\pi}$$
$$\Rightarrow \displaystyle f=\frac{h}{4\pi^2mv^2}$$
$$=\displaystyle\frac{6.6\times 10^{-34}}{4\times (3.14)^2\times 9.1\times 10^{-31}\times (0.33\times 10^{-10})^2}$$.

The distance of closest approach of a particle is given by,

For hydrogen like atom, the radius of nth orbit
$$r^z_n=\displaystyle\frac{n^2}{Z}a_0$$
Here, $$a_0=0.51\times 10^{-10}$$m
$$\therefore r^z_n=\displaystyle\frac{0.51\times 10^{-10}}{4}$$m
In the ground state, $$n=1$$
$$\therefore \displaystyle\frac{0.51\times 10^{-10}}{4}=\frac{1^2}{z}\times 0.51\times 10^{-10}$$
$$\therefore Z=4$$
So, the atom is triply ionised beryllium$$(Be^{3+})$$.

$$\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon  }_{ 0 } } \left( \cfrac { 1 }{ { r }^{ 3 } } -\cfrac { \beta  }{ { r }^{ 3 } }  \right) $$
$$\quad \therefore { v }^{ 2 }=\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon  }_{ 0 }mr } \left( \cfrac { 1 }{ { r }^{  } } +\cfrac { \beta  }{ { r }^{ 2 } }  \right) $$
From Bohr's hypothesis $$\quad v=\cfrac { nh }{ 2\pi mr } \quad \therefore \cfrac { { n }^{ 2 }{ hn }^{ 2 } }{ 4{ \pi  }^{ 2 }{ m }^{ 2 }{ r }^{ 2 } } =\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon  }_{ 0 }mr } \left( \cfrac { 1 }{ { r }^{  } } +\cfrac { \beta  }{ { r }^{ 2 } }  \right) $$
$$\therefore r+\beta ={ a }_{ 0 }{ n }^{ 2 }\quad $$

$$qvB=\cfrac { m{ v }^{ 2 } }{ r } ;mvr=\cfrac { nh }{ 2\pi  } $$
$$\therefore \cfrac { qBr }{ m } =\cfrac { nh }{ 2\pi mr } $$
$$\therefore r\propto { n }^{ 1/2 }\quad $$

$$\textbf{Correct option : option A and D}$$

$$\textbf{Explanation:}$$

We know $$T\propto n^3$$ thus ,

  $$\dfrac{T_1}{T_2} \propto (\dfrac{n_1}{n_2})^3 $$

  $$\dfrac{8}{1} \propto (\dfrac{n_1}{n_2})^3$$

  (because here $$\dfrac{T_1}{T_2} $$ is $$\dfrac{8}{1} $$)

  $$\Rightarrow \dfrac{n_1}{n_2} = \dfrac{2}{1}$$

  thus the options in which $$n_1=2n_2$$ are the correct options.

$$\textbf{Thus Option A  and option D is correct .}$$

 

$$V_{rms}=\sqrt{\cfrac{3RT}{M}}$$

$$V_{rms}$$ for helium $$=\sqrt{\cfrac{3RT}{4}}$$

$$V_{rms}$$ for argon $$=\sqrt{\cfrac{3RT}{40}}$$

$$\cfrac{V_{rms}(helium)}{V_{rms}(Argon)}=\sqrt{10}$$

$$=3.16$$

$$Given:$$ In certain electronic transition from quantum level n to ground state in atomic hydrogen in one or more steps no line belonging to Brackett series is observed.

According to Bohr's quantization rule

$$ \begin{array}{l} \text { we know, } \\ \text { Angular momentum }=\frac{n h}{2 \pi} \end{array} $$