MCQ Questions for Class 12 Medical Physics Atoms Quiz 4

Continuous spectrum is not due to

0%
Hydrogen flame
0%
Electric bulb
0%
Kerosene oil lamp flame
0%
Candle flame

Which of the following subshells is represented by the quantum numbers

$n$ =4 and

$l$ =1?

0%
$4s$
0%
$4f$
0%
$4d$
0%
$4p$

Explanation

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.

Match the following :

Type I	Type II
a) Continuous emission spectrum	e) Tungsten filament of bulb
b) Line emission spectrum	f) $CO_2$ gas
c) Band emission spectrum	g) Sodium vapour lamp
d) Line absorption spectrum	h) chromosphere of sun

0%
a-e; b-f; c-g; d-h
0%
a-f; b-e; c-h; d-e
0%
a-e; b-g; c-f; d-h
0%
a-h; b-g; c-f; d-e

Explanation

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.

A cathode ray tube has a potential difference of

$V$ between the cathode and anode. The speed of the cathode rays is given by

0%
$v \propto V$
0%
$v \propto V^{-1}$
0%
$v \propto \sqrt{V}$
0%
$v \propto V ^{2}$

Explanation

$PD = V$

$KE = eV =$

$\dfrac{1}{2} mv^2$

$m =$ mass of cathode particle

$v =$

$\sqrt{(\dfrac{2e}{m})V}$

$v =$ max velocity,

$e =$ charge on cathode particle

$v \propto \sqrt V$

Which of the following relates to photons both as wave motion and as a stream of particles?

0%
Interference
0%
$E= mc^{2}$
0%
Diffraction
0%
E $= h\nu$

Explanation

$(a)$ 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.

When the particle and its anti-particle unite, the result is

0%
A heavier particle
0%
Two or more smaller particles
0%
Photons
0%
Partly matter and partly photons

In Rutherford's

$\alpha$ -rays scattering experiment, gold foils are used because of ___________.

0%
high malleability
0%
ductility
0%
high melting point
0%
high ionisation energy

Which of the following point is not shown by the Rutherford alpha scattering experiment?

0%
$\alpha$ - particle can come within a distance of the order of $10^{-14} m$ of the nucleus
0%
The radius of the nucleus is less than $10^{-14} m$
0%
Sattering follows coulombs law
0%
The positively charged parts of an atom move with extremely high velocities

Cathode rays are made to pass between the poles of a magnet as shown in figure. The effect of magnetic field is

0%
To deflect them towards the south pole
0%
To deflect them perpendicular to the plane of the paper and towards the observer
0%
To deflect them towards the north pole
0%
To increase the velocity of the rays

The angular momentum of the

$\alpha -$ particles which are scattered through large angles by the heavier nuclei, is conserved because

0%
of the nature of repulsive forces
0%
the kinetic energy is conserved
0%
the potential energy is conserved
0%
there is no external torque

Explanation

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

The force experienced by the cathode rays when they pass through a uniform electric field of intensity

$\bar{E}$ is:

0%
in the direction of the electric field
0%
in the direction opposite to that of the electric field
0%
at right angles to the electric field
0%
zero, because cathode rays do not have any charge

Explanation

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, a screen is used to detect the alpha particles which is coated by:

0%
carbon black
0%
platinum black
0%
zinc sulphide
0%
poly tetrafluoro ethylene

An electron makes transition from

$n= 3 , n=1$ state in a hydrogen atom. The different possible number of photons that can be emitted is :

0%
$1$
0%
$2$
0%
$3$
0%
$6$

Explanation

For transition from

$n=3$ to

$n=1$
Possible number of photons

$=\ ^3C_2$

$=\dfrac{3!}{(3-2)!2!}$

$=\dfrac{3\times 2\times 1}{1\times 2\times 1}=3$

In the lowest energy level of hydrogen atom, electron has an angular momentum equal to:

0%
$\dfrac{\pi }{h}$
0%
$\dfrac{h}{\pi }$
0%
$\dfrac{h}{2\pi }$
0%
$\dfrac{2\pi }{h}$

Explanation

The angular momentum

$L$

$=m_{e}vr$ is on integer multiple of

$\dfrac{h}{2\pi }$

$mvr=\dfrac{nh}{2\pi}$

For,

$n=1$

$mvr=\dfrac{h}{2\pi}$

The ionization energy of a hydrogen-like ion

$A$ is greater than that of another hydrogen-like ion

$B$ . Let

$u$ and

$E$ represent the speed of the electron and energy of the atom respectively in the ground state. Then,

0%
$r_{A}> r_{B}$
0%
$u_{A}> u_{B}$
0%
$E_{A}> E_{B}$
0%
$L_{A}> L_{B}$

Explanation

We know

$E=-13.6\dfrac{Z^{2}}{n^{2}}eV$

So Ionization energy

$=13.6\dfrac{Z^{2}}{n^{2}}eV$

given,

$IE_{A}> IE_{B},$ so

$E_{A}< E_{B}$

that means

$Z_{A}> Z_{B}$

Now

$u \propto \dfrac{Z}{n}$

$u_{A}> u_{B}$

In the Geiger-Marsden experiment, the force that scatters particles is

0%
nuclear force
0%
coulomb force
0%
both A and B
0%
gravitational force

In Rutherford's experiment the number of

$\alpha$ particles scattered through an angle

$60^o$ is

$112$ per minute, then the number of

$\alpha$ particles scattered through an angle of

$90^{0}$ per minute by the same nucleus is:

0%
$28$ per minute
0%
$112$ per minute
0%
$12.5$ per minute
0%
$7$ per minute

Explanation

In Rutherford's experiment the number of particles scattered through an angle

$\theta$ per minute is given by:

$N \propto \dfrac{Z^2}{Sin^4(\theta /2)}$

According to the question: the number of particles scattered through an angle

$60^o$ is 112 per minute. So,

$112 \propto \dfrac{Z^2}{Sin^4(60 /2)}$ ------ (1)

And, lets assume the number of particles scattered through an angle

$90^o$ is N' per minute

$N' \propto \dfrac{Z^2}{Sin^4(90 /2)}$ ------(2)

Using (1) and (2),

$\dfrac{N'}{112} \propto \dfrac{Sin^4(60 /2)}{Sin^4(90 /2)}$

$N'=28$

The radius of hydrogen atom, when it is in its second excited state ,becomes

$\underline{\hspace{0.5in}}$ its ground state radius.

0%
half
0%
double
0%
four times
0%
nine times

Explanation

$r=\dfrac{a_{0}n^{2}}{Z}$

$r_{1}$ ( ground state )

$={a_{0}}(1)^{2}$

$=a_{0}$

$r_{2}$ ( 2nd excited state )

$=a_{0}(3)^{2}$

$=9a_{0}$

The main defect of Bohr's atom model is :

0%
mixing of classical and quantum theories
0%
exclusion of nuclear motion
0%
failed to explain the fine structure of spectral lines
0%
failed to explain larger atoms

The electron is present in an orbit of energy state

$-1.51$ eV, then angular momentum of the electron is

0%
$2h/\pi$
0%
$h/\pi$
0%
$3h/2\pi$
0%
$7h/\pi$

Explanation

Given :

$E_n =-1.51$ eV

Using

$E_n = \dfrac{-13.6}{n^2}$ eV (for H-atom)

$\therefore$

$-1.51 = \dfrac{-13.6}{n^2}$

$\implies n =3$

Thus angular momentum

$L = \dfrac{nh}{2\pi} = \dfrac{3h}{2\pi}$

The possible values of principal quantum number can be:

0%
$1, 2, 3...8$
0%
$0, 1, 2...8$
0%
only zero
0%
only odd numbers

Explanation

Values of Principle quantum number are

$1,2,3,4.....8$ .

$0$ is not a Principle quantum number.

Only odd numbers are not Principle quantum numbers either.

Odd numbers, as well as even numbers, are Principle quantum number except

$0$ and negative integers.

Atomic hydrogen is excited to the nth energy level.The maximum number of spectral lines which it can emit while returing to the ground state, is

0%
$\frac{1}{2}n(n-1)$
0%
$\frac{1}{2}n(n+1)$
0%
$n(n+1)$
0%
$n(n-1)$

Explanation

Maximum no of spectral lines

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

$A_{n}$ is the area enclosed in the nth orbit in a hydrogen atom then the graph log

$\left ( \dfrac{A_{n}}{A_{1}} \right )$ against log

$n$

0%
will have slope $2$ (straightline)
0%
will have slope $4$ (straightline)
0%
will be a monotonically increasing non linear curve
0%
will be a circle

Explanation

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

$=\pi a{_{0}}^{2}A_{1}$

$\dfrac{A_{n}}{A_{1}}=n^{4}\left [ Taking\ log\ both\ sides \right ]$

$\Rightarrow log\left ( \dfrac{A_{n}}{A_{1}} \right )=4logn$

Slope of the above line

$=4$

If an electron is revolving round the hydrogen nucleus at a distance of

$0.1\ nm$ , the speed should be :

0%
$2.188\times 10^{6}m/s$
0%
$1.094\times 10^{6}m/s$
0%
$4.376\times 10^{6}m/s$
0%
$1.59\times 10^{6}m/s$

Explanation

Radius of orbit

$=0.1nm$
According to formula:

$0.1\times 10^{-9}=0.529\times 10^{-10}\times n^{2}$

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

So, in 2nd case:
Velocity of electron

$=\dfrac{2.18\times 10^{6}}{n}$

$\Rightarrow V=\dfrac{2.18}{1.37}\times 10^{6}=1.59\times 10^{6}m/s$

When an electron jumps from higher orbit to the second orbit in hydrogen, the radiation emitted out will be in

$(R=1.09\times 10^{7}m^{-1})$

0%
ultraviolet
0%
visible region
0%
infrared region
0%
X-ray region

The maximum number of photons emitted by an H-atom, if atom is excited to states with principal quantum number four is

0%
$4$
0%
$6$
0%
$2$
0%
$1$

Explanation

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.

The velocity of a helium nucleus travelling in a curved path in a magnetic field is

$V$ . The velocity of a proton moving in the same curved path in the same magnetic field is :

0%
$V$
0%
$4V$
0%
$2V$
0%
$V/2$

Explanation

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

$= 2V$

There are only three hydrogen atoms in a discharge tube. The analysis of spectrum shows that in all the hydrogen atoms, electrons are de-exciting from the fourth orbit. What should be the maximum number of spectral lines?

0%
$6$
0%
$1$
0%
$4$
0%
$5$

Explanation

Since, from the

$n^{th}$ state, the electron may go to

$(n - 1)^{th}$ state,

$(n - 2)^{th}$ state ,... 2nd state or 1st state. So there are (n 1) possible transitions starting from the

$n^{th}$ state. The atoms reaching

$(n - 1)^{th}$ state may make (n 2) different transitions. The atoms reaching

$(n - 2)^{th}$ state may make (n 3) different transitions. Similarly for other lower states. During each transition a photon with energy

$h\nu$ and wavelength

$\lambda$ is emitted out. Hence, the total number of possible transitions is equal to the number of photons emitted.

The total number of possible transitions are

$(n-1), (n-2), (n-3), ................, 3, 2, 1 = \dfrac{n(n-1)}{2}$

Therefore, for transition of an electron from higher energy state

$n = 4$ to lower energy state

$n = 1$ the number of photons emitted are

$\dfrac{n(n-1)}{2} = \dfrac{4(4-1)}{2} = \dfrac{12}{2} = 6$

Here, the electrons of each hydrogen atom are in same state hence in transition will emit photons of equal wavelengths for each transition hence, the maximum spectral lines emitted are

$6$ .

Rydberg atoms are the hydrogen atoms in higher excited states such atoms are observed in space.The orbit number for such an atom with radius about

$0.01\ mm$ should be :

0%
$1$
0%
$435$
0%
$13749$
0%
$117$

Explanation

Given the radius of the orbit of the exited electron

$=0.01mm$
According to formula:

$\Rightarrow$ radius

$=0.529A^{0}(n^{2})$

$\Rightarrow 0.01\times 10^{-3}=0.529\times 10^{-10}\times n^{2}$

$\Rightarrow \dfrac{0.01\times 10^{-3}}{0.529\times 10^{10}}=n^{2}\Rightarrow n^{2}=189035.9168$

$\Rightarrow n=434.7826\Rightarrow n\approx 435$

The radius of shortest orbit in one electron system is

$18$ pm.It may be.

0%
$_{1}^{1}H$
0%
$_{2}^{1}H$
0%
$He^{+}$
0%
$Li^{+}$

Explanation

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.

The threshold wavelength for a surface having a threshold frequency of

$0.6\times 10^{15}$ Hz in (

$\mathring A$ ) is

0%
$100$
0%
$2000$
0%
$5000$
0%
$400$

Explanation

$\begin{array}\text { given, } \\\text { threshold wavelength, } \lambda_{0} =\frac{c}{f}\\c=\text { Speed of light }=3 \times 10^{8}\mathrm{~ms} \\f =\text { frequency }=0.6\times 10^{15} \\=6 \times 10^{14}\mathrm{~Hz}\end{array}$

$\begin{aligned}\therefore \quad \lambda_{0}&=\frac{3 \times 10^{8}}{6 \times10^{14}} \mathrm{~m} \\&=5 \times 10^{-7} \mathrm{~m} \\&=5000 \times 10^{-10} \mathrm{~m} \\&=5000 \dot A\end{aligned}$

In one revolution round the hydrogen nucleus, an electron makes five crests .The electron belongs to

0%
$1^{st}$ orbit
0%
$4^{th}$ orbit
0%
$5^{th}$ orbit
0%
$6^{th}$ orbit

Explanation

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.

The ratio of momenta of an electron and a $\alpha$ -particle which is accelerated from rest by a potential difference of $100 V$ is:

0%
1
0%
$\displaystyle \sqrt {\dfrac{{2{m_e}}}{{{m_\alpha }}}}$
0%
$\displaystyle \sqrt {\dfrac{{{m_e}}}{{{m_\alpha }}}}$
0%
$\displaystyle \sqrt {\dfrac{{{m_e}}}{{2{m_\alpha }}}}$

Explanation

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

where,

$Me = mass \ of \ electron$

$Ma = mass \ of \ alpha \ particle$

$Qe = charge \ of \ electron$

$Qa = charge \ of \ alpha \ particle$

we know,

$Qa = 2Qe$

therefore

$\dfrac{Pe}{Pa}=\sqrt(\dfrac{MeQe}{2MaQe})$

So we get,

$\dfrac{Pe}{Pa}=\sqrt(\dfrac{Me}{2Ma})$ .

The energy of a hydrogen atom in the ground state is 13.6 eV. The energy of He+ ion in the first excited state will be

0%
-13.6 eV
0%
-27.2 eV
0%
-54.4 eV
0%
-6.8 eV

According to Bohr's theory of hydrogen atom, for an electron in the

$n^{th}$ allowed orbit, then

0%
linear momentum is proportional to $(1/n)$
0%
radius is proportional to $n$
0%
the kinetic energy is proportional to $(1/n^2)$
0%
the angular momentum is proportional to $n$

Explanation

Linear momentum is inversely proportional to

$n$ and angular momentum is directly proportional to

$n$ .

We have the equation,

$E=-13.6\cfrac{z^2}{n^2}$

So,

$K.E. \alpha \cfrac{1}{ n^2 }$

When a hydrogen atom is in its first excited level, its radius is

0%
four times its ground state radius
0%
twice times its ground state radius
0%
same times its ground state radius
0%
half times its ground state radius.

Hydrogen atom is excited from ground state to another state with principal quantum number equal to

$4$ . Then the number of spectral lines in the emission spectra will be.

0%
$2$
0%
$3$
0%
$5$
0%
$6$

Explanation

Total number of spectral lines,

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

here,

$n$ is principal quantum number, given

$n = 4$

$\Rightarrow N = \dfrac {4(4 - 1)}{2} = 6$

Hence, the number of spectra lines in the emission spectrum will be

$6$ .

Gold is chosen by Rutherford for his

$\alpha$ -ray scattering experiment because:

0%
gold has high malleability
0%
gold has high ductility
0%
gold has high density
0%
gold is the least reactive element

Rutherford's alpha (

$\alpha$ ) particle scattering experiment resulted in discovery of:

0%
electron
0%
proton
0%
nucleus in the atom
0%
atomic mass

Explanation

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.

On the basis of his experiment,
Rutherford put forward the nuclear model of an atom, which had the following features:
(i) There is a positively charged center in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.
(ii) The electrons revolve around the nucleus in well-defined orbits.
(iii) The size of the nucleus is very small as compared to the size of the atom. Rutherford's alpha particle scattering experiment shows the presence of nucleus in the atom.
It also gives the following important information about the nucleus of an atom :
(i) The nucleus of an atom is positively charged.
(ii) The nucleus of an atom is very dense and hard.
(iii) The nucleus of an atom is very small as compared to the size of the atom as a whole.
Rutherford model of the atom is also called Nuclear model of the atom.

Hence (C) is the correct option.

The time taken by a photo-electron to come out after the photon strikes is approximately

0%
$10^{-6}$ sec
0%
$10^{-4}$ sec
0%
$10^{-10}$ sec
0%
$10^{-16}$ sec

Explanation

The time by a photoeletron to come out after the
photon strikes is approximately

$10^{-10}$ seconds. This is a fact.

So, the answer is option (C).

Rutherford's model explains :

0%
Atom is a planetary model.
0%
Electrons never loose nor gain energy.
0%
Electron revolves around the nucleus with high velocities to counterbalance the forces of electrostatic forces of attraction between protons and electrons.
0%
Electrons do not move at all.

When a hydrogen atom emits a photon of energy

$12.1 eV$ , its orbital angular momentum changes by

0%
$1.05 \times 10^{-34}$ Js
0%
$2.11 \times 10^{-34}$ Js
0%
$3.16 \times 10^{-34}$ Js
0%
$4.22 \times 10^{-34}$ Js

Explanation

When a hydrogen atom emits a photon of energy 12.1 eV, i.e. the energy difference between two levels hence,

$E_{n_2} - E_{n_1} = 12.1$

$E_{n_2} = 12.1 + E_{n_1}$

$E_{n_2} = 12.1 + (-13.6)$

$E_{n_2} = -1.5 eV$

This energy corresponds to the third orbit hence, change in orbital momentum is:

$\Delta L = \dfrac{h}{2\pi} (n_2 - n_1)$

$\Delta L = \dfrac{h}{2\pi} (3 - 1) = \dfrac{h}{\pi}$

$\Delta L = \dfrac{6.6 \times 10^{-34} }{3.14}$

$\Delta L = 2.11 \times 10^{-34} Js$

The fine structure of hydrogen spectrum can be explained by

0%
the presence of neutrons in the nucleus.
0%
the finite size of nucleus.
0%
the orbital angular momentum of electrons.
0%
the spin angular momentum of electrons.

In the hydrogen atom in the ground state

0%
the kinetic energy of the electron is less than the potential energy which is positive
0%
the potential energy is less than the kinetic energy which is positive
0%
the potential energy is negative and the kinetic energy numerically less than the numerical value of potential energy
0%
the total energy is negative

Explanation

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

$\begin{array}{l} \text { Total Energy= Kinetic Energy+Potential energy } \\ \qquad \begin{array}{l}T \cdot E= K \cdot E +P \cdot E \end{array} \\ \text { & } K \cdot E = - \frac{P \cdot E}{2} \end{array}$

$\begin{aligned} \therefore T \cdot E =\frac{-P \cdot E \cdot}{2}+P \cdot E \cdot \\ \frac{-13 \cdot 6 z^{2}}{n^{2}} e v &=\frac{P \cdot E}{2} \end{aligned}$

$\begin{aligned} \therefore P \cdot E \cdot &=-\frac{2 \times 13 \cdot 6 \times z^{2}}{n^{2}} \\ &=-27 \cdot 2 \times \frac{z^{2}}{n^{2}} \end{aligned}$

$\begin{aligned} \text { & } K \cdot E \cdot &=-\frac{P \cdot E}{2} \\ &=13 \cdot 6 \frac{z^{2}}{n^{2}} e v \\ & \text { for hydrogen's ground state } \\ z=1, & n=1 \end{aligned}$

$\begin{aligned} P \cdot E =& \frac{-27 \cdot 2 \times (1)^{2}}{1} \\ =&-27. 2 e v \\ \& K.E= \frac{13.6 \times (1)^2}{1} = 13.6ev \\ \text { Hence, option } c \text { is correct. } \end{aligned}$

Rutherford experiment of scattering of

$\alpha$ particles showed for the first time that the atom has:

0%
electrons
0%
only protons
0%
nucleus
0%
None of the above

Explanation

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.
Draw a labelled diagram for α-particle scattering experiment. Give Rutherford s observation and discuss the significance of this experiment. from Physics Atoms Class 12 CBSE

Final answer: The correct answer is $C$ .

A small positively charged nucleus is present in the center. Which observations tells about this?

0%
Most of the $\alpha \ -$ particles passed straight
0%
Most of the $\alpha \ -$ particles rebounded after hitting the atoms
0%
Only a few $\alpha \ -$ particles deflected away from their path
0%
Very few $\alpha \ -$ particles rebounded

Explanation

The observations of

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

Hence,
(i) Only a few

$\alpha$ - particles deflected away from their path.
(ii) Very few

$\alpha$ - particles rebounded.

Which of the following products in a hydrogen atom is independent of principal quantum number

$n$ ?

0%
$v_r$
0%
$v_n$
0%
$E_r^{2}$
0%
$E_n$

Explanation

The velocity of revolving electron is:

$v = \dfrac{e^2}{2nh\epsilon_0}$

Hence,

$v_n = \dfrac{e^2}{2nh\epsilon_0}(n)$

$v_n = \dfrac{e^2}{2h\epsilon_0}$

The product thus obtained does not contain principal quantum number(n) hence

$v_n$ is independent of

$n$ .

When a photomultiplier tube was used , the photo current recorded is 60

$\mu A$ . The actual photo current is

0%
$\displaystyle > 60$ $\mu A$
0%
$\displaystyle = 60$ $\mu A$
0%
$\displaystyle < 60$ $\mu A$
0%
None of these

Explanation

The current has to be

$< 60 \mu A$ .
Since dark current is also involved in the measured current value,

$I_{actual} = I_{measured} - I_{dark}$

Cathode ray oscillograph is used for

0%
Taking X-ray photographs
0%
Showing pictures
0%
Demonstrating electrical pulses
0%
Producing elementary particles

Explanation

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

Rutherford's scattering experiment is related to the size of the:

0%
nucleus
0%
atom
0%
electron
0%
neutron

Explanation

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.

CBSE Questions for Class 12 Medical Physics Atoms Quiz 4 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 4 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.