Hydrogen is molecular, therefore it gives band spectrum but not continuous spectrum.

The principle quantum number $$n = 4$$ , represents the fourth orbit. A subshell is the set of states defined by a common azimuthal quantum number l, within a shell. The values $$l = 0, 1, 2, 3$$ correspond to the s, p, d, and f shells, respectively. Hence, $$n = 4$$ , $$l = 1$$ represents 4p subshell.

Continuous emission spectrum can be seen for Tungsten emission filament of bulb.
Line emission spectrum is observed when sodium Vapour lamp is used.
Band emission spectrum is observed for molecules like $$CO_{2}$$ gas.
Line absorption spectrum is observed when elements in chromospere of sun absorb light of their characteristics wavelength.

$$PD = V$$
$$KE = eV =$$ $$\dfrac{1}{2} mv^2$$     $$m =$$ mass of cathode particle
$$v =$$ $$\sqrt{(\dfrac{2e}{m})V}$$       $$v =$$ max velocity,  $$e =$$ charge on cathode particle
$$v \propto \sqrt V$$

$$(a)$$ Interference exhibits the wave nature of light.
$$(b)$$ $$E = m c^{2}$$ exhibits the particles nature of light
$$(c)$$ Diffraction exhibits the wave nature of light.
$$(d)$$ $$E = h \nu$$ shows both particle nature and wave nature of light, i.e., light is having a particular wavelength $$\lambda$$ and the energy is in the form of packets.

Malleability is the material's ability to form thin sheets. In Rutherford's ray scattering experiment, gold foil is used because of its high malleability. Very thin gold foil is used in the experiment and gold is capable of being rolled into extremely thin foils.

The conclusions of Rutherford's alpha scattering experiment do not include that the positively charged particles move with great velocities. The positively charged particles are present at the center of the atom called a nucleus.

Force will be perpendicular and outside to the plane of the paper

The angular momentum of $$\alpha$$ particles is conserved because there is no external torque.

Cathode particles are negatively charged
$$\therefore F_E = -q \vec{E}$$
Force is opposite to the direction of electric field

In Rutherford's alpha-ray scattering experiment, the alpha particles are detected using a screen coated with zinc sulphide.

For transition from $$n=3$$ to $$n=1$$
Possible number of photons $$=\ ^3C_2$$ $$=\dfrac{3!}{(3-2)!2!}$$ $$=\dfrac{3\times 2\times 1}{1\times 2\times 1}=3$$

The angular momentum $$L$$ $$=m_{e}vr$$ is on integer multiple of $$\dfrac{h}{2\pi }$$

$$mvr=\dfrac{nh}{2\pi}$$  

We know

Coulomb force is responsible for scattering of particles. When the alpha particles (positive in charge) get closer to the nucleus, which is positive in charge, they get repelled through various angles.

In Rutherford's experiment the number of  particles scattered through an angle $$\theta$$ per minute is given by:
$$N \propto \dfrac{Z^2}{Sin^4(\theta /2)}$$

$$r=\dfrac{a_{0}n^{2}}{Z}$$
$$r_{1}$$ ( ground state ) $$={a_{0}}(1)^{2}$$
                                $$=a_{0}$$
$$r_{2}$$ ( 2nd excited state ) $$=a_{0}(3)^{2}$$
                                        $$=9a_{0}$$

Bohr's model can only clearly explain hydrogen or hydrogen-like atoms, it fails when applied to larger and heavier atoms like iron, gold, mercury, etc.

Given :     $$E_n =-1.51$$ eV

Values of Principle quantum number are $$1,2,3,4.....8$$ . $$0$$ is not a Principle quantum number. 

Maximum no of spectral lines $$=\dfrac{n(n-1)}{2}$$

Radius of $$n^{th}$$ orbit $$=a_{0}n^{2}$$
Radius of 1st orbit $$=a_{0}$$
Area of $$n^{th}$$ orbit $$=\pi a{_{0}}^{2}n^{4}=A_{n}$$
Area of 1st orbit $$=\pi a{_{0}}^{2}A_{1}$$
$$\dfrac{A_{n}}{A_{1}}=n^{4}\left [ Taking\ log\ both\ sides \right ]$$
$$\Rightarrow log\left ( \dfrac{A_{n}}{A_{1}} \right )=4logn$$

Radius of orbit $$=0.1nm$$
According to formula:
$$0.1\times 10^{-9}=0.529\times 10^{-10}\times n^{2}$$

$$\Rightarrow n^{2}=1.8\Rightarrow n=1.37$$

Hint:

Number of Photons emitted by H-atom if it is excited to  $${n^{th}}$$ excited state is,

$$\bf{photons = \dfrac{{n(n - 1)}}{2}}$$

                                                                           

Step 1: Calculate the number of photons emitted by H-atom

Hydrogen atom is excited to $${4^{th}}$$ excited state, so number of photons emitted by using formula is,

$$photons = \dfrac{{n(n - 1)}}{2}$$

$$\Rightarrow photons = \dfrac{{4(4 - 1)}}{2}$$

$$\Rightarrow photons = 6$$

Thus, 6 photons are emitted if the Hydrogen atom is excited to $${4^{th}}$$ excited state.

Option B is correct.

Charge on helium nucleus $$=2\alpha$$

Mass of a helium nucleus $$= 4\beta$$

Where $$\alpha$$ and $$\beta$$ are charge and mass of the helium nuclei respectively.

According to question,

$$V = \dfrac{Bqr} {m} = \dfrac{2B\alpha r} {4\beta}$$ ------------------- (I)

For proton:
Velocity $$=\dfrac{B\alpha r} {\beta} = 2V$$                 [From (I)]

Velocity of proton $$= 2V$$  

Given the radius of the orbit of the exited electron $$=0.01mm$$
According to formula:
$$\Rightarrow$$ radius $$=0.529A^{0}(n^{2})$$
$$\Rightarrow 0.01\times 10^{-3}=0.529\times 10^{-10}\times n^{2}$$
$$\Rightarrow \dfrac{0.01\times 10^{-3}}{0.529\times 10^{10}}=n^{2}\Rightarrow n^{2}=189035.9168$$
$$\Rightarrow n=434.7826\Rightarrow n\approx 435$$  

 

Hint:

For an atom of atomic number Z, The radius of $${n^{th}}$$  orbit is given as,

$${r_n} = \dfrac{{{n^2}{a_o}}}{Z}$$

Where, $${a_o}$$ is Bohr’s radius, $${a_o} = 53pm$$

 

Correct Option : Option D.

Explanation for correct answer:

a)      $${_1^1}H$$

$$Z = 1$$

For shortest orbit, $$n = 1$$

Therefore, $${r_1} = \dfrac{{{1^2} \times 53}}{1} = 53pm$$

 

b)      $${_2^1}H$$

$$Z = 1$$

For shortest orbit, $$n = 1$$

Therefore, $${r_1} = \dfrac{{{1^2} \times 53}}{1} = 53pm$$

 

c)      $$H{e^ + }$$

$$Z = 2$$

For shortest orbit, $$n = 1$$

Therefore, $${r_1} = \dfrac{{{1^2} \times 53}}{2} = 26.5pm$$

 

d)      $$L{i^ + }$$

$$Z = 3$$

For shortest orbit, $$n = 1$$

Therefore, $${r_1} = \dfrac{{{1^2} \times 53}}{3} = 17.667pm$$

The radius of shortest orbit of $$L{i^ + }$$ is closest to $$18pm$$ . Thus Option D is the correct answer.

As the interaction of wavelength must be constructive in interference.  So the wavelength must be quantised.
So, $$n_{0}\lambda =2\pi r$$
where $$n$$ is the quantum state or the number of orbit i.e., 5 in this case, thus 5 walengths is the circumference of the orbit with 5 crests.  

$$\dfrac{Pe}{Pa}=\sqrt(\dfrac{MeQe}{MaQa})$$

hint: using the radius formula of Bohr model

Total number of spectral lines,

Malleability is a property of metals which can be beaten into sheets. For his experiment Rutherford needed a very thin sheet of metal and hence he selected gold metal which could be beaten into very thin sheet.

Rutherford concluded from the $$\alpha$$ -particle scattering experiment that:
(i) Most of the space inside the atom is empty because most of the  $$\alpha$$ -particles passed through the gold foil without getting deflected. 
(ii) Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space. 
(iii) A very small fraction of  $$\alpha$$ -particles were deflected by very large angles, indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom. 

The time by a photoeletron to come out after the
photon strikes is approximately $$10^{-10}$$  seconds. This is a fact.

According to Rutherford, 

When a hydrogen atom emits a photon of energy 12.1 eV, i.e. the energy difference between two levels hence,
$$E_{n_2} - E_{n_1} = 12.1$$
$$E_{n_2} = 12.1 + E_{n_1}$$
$$E_{n_2} = 12.1 + (-13.6)$$
$$E_{n_2} = -1.5 eV$$

$$\begin{array}{l} \text { we know that total energy } \\ \text { of an electron in } n^{\text {th }} \text { orbit } \\ \text { is }=-13.6 \frac{z^{2}}{n^{2}} ev \\ \text { Total Energy }=-13.6 \frac{z^{2}}{n^{2}} e v \end{array}$$  

Hint: The Rutherford alpha particle scattering experiment showed that most of the space of the atom is empty

Explanation:
The Rutherford alpha particle scattering experiment has the following observation:
 1. Most of the alpha particle remains undeflected that means most of the space was empty and the particles came undeflected by positive charges in the atom. It was a very small portion and termed as the nucleus.

2. Some of the alpha particles show a small deviation from the original path.

3. Very few particles bounced back

So, Conclusions can be made from this as;

1. Most of the space in the atom remains empty.

2. And a large amount of mass is concentrated in the small space called a nucleus.

Final answer: The correct answer is $$C$$ .

The observations of $$\alpha \ -$$ ray scattering experiment lead to the presence of small positively charged nucleus in the centre. 

The velocity of revolving electron is:
$$v = \dfrac{e^2}{2nh\epsilon_0}$$

The current has to be $$< 60 \mu A$$ .
Since dark current is also involved in the measured current value,
$$I_{actual} = I_{measured} - I_{dark}$$

$$\begin{array}{l}\text { The cathode ray oscilloscope (CRO) } \\\text { is a common laboratory instrument } \\\text { that provides accurate time } \\\text { and aplitude measurements of } \\\text { voltage signals over a wide } \\\text { range of frequencies. Its } \\\text { reliability, stability and ease of } \\\text { opteration make it suitable } \\\text { as a general purpose laboratory } \\\text { instrument. } \\\text { In easy words, it is used } \\\text { for  demonstrating } \\\text { electrical pulses. }\end{array}$$

Rutherford scattering experiment is related to size of the nucleus. Rutherford defined that an atom consists of a large space where $$\alpha -$$ particles pass through, but only at a small region they are deflected. This region is nucleus.