Discovery and Properties of anode, cathode rays neutron and Nuclear structure Questions in English

(N/A) In an evacuated cathode ray discharge tube,when a sufficiently high voltage is applied across the electrodes,current starts flowing through a stream of particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode). These were called cathode rays or cathode ray particles.
Characteristics of cathode rays or results of cathode ray discharge tube experiments:
$(i)$. The cathode rays start from the cathode and move towards the anode.
$(ii)$. These rays themselves are not visible,but their behavior can be observed with the help of certain kinds of materials (fluorescent or phosphorescent) which glow when hit by them.
$(iii)$. In the absence of an electrical or magnetic field,these rays travel in straight lines.
$(iv)$. In the presence of an electrical or magnetic field,the behavior of cathode rays is similar to that expected from negatively charged particles,suggesting that the cathode rays consist of negatively charged particles,called electrons.
$(v)$. The characteristics of cathode rays (electrons) do not depend upon the material of the electrodes and the nature of the gas present in the cathode ray tube. Thus,we can conclude that electrons are a basic constituent of all atoms.

(N/A) In $1897$,$J.J. Thomson$ determined the ratio of electrical charge $(e)$ to the mass of an electron $(m_e)$ using a cathode ray tube.
The value of the ratio is:
$e / m_e = 1.758820 \times 10^{11} \ C \ kg^{-1}$
Where:
$m_e = \text{mass of the electron in } kg$
$e = \text{magnitude of the charge on the electron in coulomb } (C)$
Method:
$Thomson$ applied electric and magnetic fields perpendicular to each other on the path of the electrons.
Observations:
$1$. When only the electric field is applied,the electrons deviate from their path and hit the fluorescent screen at point $A$.
$2$. When only the magnetic field is applied,the electrons strike the screen at point $C$.
$3$. By carefully balancing the strengths of the electric and magnetic fields,it is possible to bring the electrons back to their original path,hitting the screen at point $B$.
By performing accurate measurements of these deflections,$Thomson$ was able to determine the value of $e / m_e$.

(D) According to $J.J. Thomson$,the deflection of cathode rays (electrons) depends on the following factors:
$1$. The magnitude of the negative charge on the particle: Greater the magnitude of the charge,greater is the interaction with the electric or magnetic field,and thus greater is the deflection.
$2$. The mass of the particle: Lighter the particle,greater is the deflection.
$3$. The strength of the electric or magnetic field: The deflection increases with the increase in the voltage across the electrodes or the strength of the magnetic field.

The charge of an electron was determined by $R. A. Millikan$ using the oil drop experiment.
The charge of an electron is approximately $-1.6022 \times 10^{-19} \ C$.
The mass of an electron is calculated using the ratio of charge to mass $(e/m_e)$:
$m_e = \frac{e}{e/m_e} = \frac{1.6022 \times 10^{-19} \ C}{1.758820 \times 10^{11} \ C \ kg^{-1}} = 9.1094 \times 10^{-31} \ kg$.

(N/A) The fundamental particles of an atom are the electron,proton,and neutron. Their characteristics are summarized in the table below:

Properties	Electron	Proton	Neutron
Symbol	$e$ or ${}_{-1}e^{0}$	$p$ or ${}_{1}^{1}p$	$n$ or ${}_{0}^{1}n$
Absolute charge $(C)$	$-1.6022 \times 10^{-19}$	$+1.6022 \times 10^{-19}$	$0$
Relative charge	$-1$	$+1$	$0$
Mass $(kg)$	$9.10939 \times 10^{-31}$	$1.67262 \times 10^{-27}$	$1.67493 \times 10^{-27}$
Mass $(u)$	$0.00054$	$1.00727$	$1.00867$
Approx Mass $(u)$	$0$	$1$	$1$

(N/A) Electrical discharge carried out in a modified cathode ray tube led to the discovery of particles carrying a positive charge,also known as canal rays.
Characteristics of canal rays:
$(i)$ The positively charged particles depend upon the nature of the gas present in the cathode ray tube. These are simply the positively charged gaseous ions.
$(ii)$ The charge-to-mass ratio of the particles is found to depend on the gas from which they originate.
$(iii)$ Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
$(iv)$ The behavior of these particles in a magnetic or electrical field is opposite to that observed for electrons or cathode rays.
The smallest and lightest positive ion was obtained from hydrogen and was called a proton $(H^{+})$.

(N/A) Electrical discharge carried out in a modified cathode ray tube led to the discovery of particles carrying a positive charge,which are known as canal rays.
Characteristics of canal rays:
$(i)$ The positively charged particles depend upon the nature of the gas present in the cathode ray tube. These are simply positively charged gaseous ions.
$(ii)$ The charge-to-mass ratio of the particles depends on the gas from which they originate.
$(iii)$ Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
$(iv)$ The behavior of these particles in a magnetic or electrical field is opposite to that observed for electrons or cathode rays.
The smallest and lightest positive ion was obtained from hydrogen and was called a proton $(H^{+})$.

(N/A) The mass of one neutron is given as $1.675 \times 10^{-27} \ kg$.
To find the mass of $1$ mole of neutrons,we multiply the mass of a single neutron by Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$.
Mass of $1$ mole = $1.675 \times 10^{-27} \ kg \times 6.022 \times 10^{23} \ mol^{-1} = 1.0088 \times 10^{-3} \ kg$.

(D) $Neutron$ is a neutral particle with no net electric charge.
Because it lacks a charge,it does not experience any force when passing through an electric field,and therefore,it will not show any deflection from its path.
In contrast,$Protons$ (positively charged),$Cathode$ $rays$ (electrons,negatively charged),and $Electrons$ (negatively charged) are charged particles and will experience a force,resulting in deflection from their path when passing through an electric field.

(B) The term $Atom$ is derived from the Greek word '$atomos$'.
It means '$indivisible$' or '$uncuttable$'.

(B) Cathode rays are streams of negatively charged particles known as $electrons$.
These rays are produced when a high voltage is applied across a discharge tube containing gas at a very low pressure.
The rays originate from the $cathode$ (negative electrode) and travel towards the $anode$ (positive electrode).

(N/A) The charge of an electron $(e)$ is $-1.6022 \times 10^{-19} \ C$.
The mass of an electron $(m_{e})$ is $9.10939 \times 10^{-31} \ kg$.
The charge-to-mass ratio is calculated as $\frac{e}{m_{e}}$.
$\frac{e}{m_{e}} = \frac{-1.6022 \times 10^{-19} \ C}{9.10939 \times 10^{-31} \ kg} = 1.758820 \times 10^{11} \ C \ kg^{-1}$.

(N/A) $J.J. \text{ Thomson}$ determined the value of $e/m_e$ by applying perpendicular electric and magnetic fields to the path of electrons in a cathode ray tube experiment.
$R.A. \text{ Millikan}$ determined the charge of an electron using the oil drop experiment.

(A) The proton was discovered by $Goldstein$ in $1886$ through his experiments with anode rays.
The neutron was discovered by $James \ Chadwick$ in $1932$ by bombarding beryllium with alpha particles.

(A) In $1932$,James Chadwick discovered the neutron by bombarding a thin sheet of beryllium with $\alpha$-particles.
This reaction produced electrically neutral particles,which he named neutrons.
The nuclear reaction is: $^9_4Be + ^4_2He \rightarrow ^{12}_6C + ^1_0n$.

(N/A) The formula for Coulombic force is $F_c = \frac{k q_1 q_2}{r^2}$,where $k$ is Coulomb's constant,$q_1$ and $q_2$ are the charges,and $r$ is the distance between them.
The formula for gravitational force is $F_g = \frac{G m_1 m_2}{r^2}$,where $G$ is the gravitational constant,$m_1$ and $m_2$ are the masses,and $r$ is the distance between them.

(N/A) No,neutrons are not present in all atomic nuclei. For example,the most common isotope of hydrogen,$^1H$ (protium),contains only one proton and no neutrons in its nucleus.

(A) The diameter of the nucleus is approximately $10^{-13} \ cm$ to $10^{-12} \ cm$.
The diameter of an atom is approximately $10^{-8} \ cm$.
Therefore,the ratio of the diameter of an atom to the diameter of the nucleus is:
$\frac{\text{Diameter of atom}}{\text{Diameter of nucleus}} = \frac{10^{-8} \ cm}{10^{-13} \ cm} = 10^5$.
Thus,the atom is $10^5$ times larger than its nucleus.

(N/A) The intensity of spectral lines depends on the number of photons emitted or absorbed during the transition of electrons between energy levels.

(B) The production of a scintillation (flash of light) on a $ZnS$ screen upon the impact of an electron indicates that the electron behaves as a discrete particle. This observation is a key piece of evidence for the particle nature of electrons.

(N/A) $(i)$ Electricity
$(ii)$ Cathode rays
$(iii)$ Zinc sulphide
$(iv)$ Cathode ray

(N/A) $(i)$ Phosphorescent material or fluorescent material
$(ii)$ Electrons
$(iii)$ $J$.$J$. Thomson
$(iv)$ Greater

(N/A) $(i)$ Oil-drop method
$(ii)$ Canal rays (or Anode rays)
$(iii)$ Proton
$(iv)$ Neutron

(N/A) $(i)$ Beryllium
$(ii)$ $100 \ nm$
$(iii)$ $10^{-10} \ m, 10^{-15} \ m$
$(iv)$ Nucleus

(A) $(i)$ Orbit (or Shell)
$(ii)$ Electrostatic force of attraction
$(iii)$ $Z$ and $A$
$(iv)$ Nucleons and $A$

(A) $(i)$ False. Faraday's experiments and cathode ray tubes do not directly prove the location of protons in the nucleus; this was established by Rutherford's alpha-particle scattering experiment.
$(ii)$ False. Goldstein discovered anode rays (canal rays) in $1886$,which were later identified as protons,but the discovery is generally attributed to the $1886$ experiments.
$(iii)$ True. James Chadwick discovered the neutron in $1932$.
$(iv)$ True. $J$.$J$. Thomson proposed the first atomic model (Plum Pudding Model) in $1898$.

(B) In atomic structure,the term $ \text{kernel} $ refers to the nucleus of an atom along with all the electrons present in the inner shells,excluding the valence electrons.
Essentially,$ \text{Kernel} = \text{Nucleus} + \text{Inner shell electrons} $.
It represents the core of the atom that remains relatively stable during chemical bonding.

(C) The symbol $\alpha$ represents an alpha particle.
An alpha particle consists of $2$ protons and $2$ neutrons,which is identical to the nucleus of a helium atom $(^4_2He^{2+})$.
Therefore,it is referred to as a helium nucleus.

(B) .
Cathode rays are observed only at low pressure and high voltage.
In the absence of external electrical or magnetic fields,they travel in straight lines.
The characteristics of cathode rays are independent of the material of the electrodes or the nature of the gas present in the tube.

(A) Calculation of de Broglie wavelength: $\lambda = \frac{h}{mv} = \frac{6 \times 10^{-34} \, Js}{(9 \times 10^{-31} \, kg) \times (1000 \, m \, s^{-1})} = 6.6667 \times 10^{-7} \, m = 666.67 \, nm$. Thus,statement $(A)$ is true.
$(B)$ The characteristics of cathode rays (electrons) are independent of the material of the electrodes. Thus,statement $(B)$ is false.
$(C)$ Cathode rays originate from the cathode and travel towards the anode. Thus,statement $(C)$ is true.
$(D)$ The nature of cathode rays is independent of the nature of the gas present in the tube. Thus,statement $(D)$ is false.
Therefore,there are $2$ true statements ($A$ and $C$).

(D) The $e/m$ ratio is defined as the charge-to-mass ratio of a particle.
For a neutron $(n)$,the charge is $0$,so $e/m = 0$.
For an alpha particle $(\alpha)$,the charge is $+2$ and the mass is $4 \ amu$,so $e/m = 2/4 = 0.5$.
For a proton $(p)$,the charge is $+1$ and the mass is $1 \ amu$,so $e/m = 1/1 = 1$.
For an electron $(e^-)$,the charge is $-1$ and the mass is $1/1837 \ amu$,so $e/m = 1 / (1/1837) = 1837$.
Comparing these values: $0 < 0.5 < 1 < 1837$.
Thus,the increasing order is $n < \alpha < p < e$.

(D) Characteristics of cathode rays are:
$(1)$ The cathode rays start from the cathode and move towards the anode. Thus,statement $(1)$ is incorrect.
$(2)$ These rays themselves are not visible,but their behavior can be observed with the help of fluorescent materials such as $ZnS$. Thus,statement $(2)$ is incorrect.
$(3)$ In the presence of electrical or magnetic fields,cathode rays behave like negatively charged particles,suggesting that they consist of electrons. Thus,statement $(3)$ is incorrect.
$(4)$ The characteristics of cathode rays (electrons) do not depend upon the material of the electrodes or the nature of the gas present in the cathode ray tube. Hence,statement $(4)$ is correct.

(A) The mass of an electron is $m_e = 9.109 \times 10^{-31} \ kg$.
The mass of a proton is $m_p = 1.672 \times 10^{-27} \ kg$.
The mass of a neutron is $m_n = 1.674 \times 10^{-27} \ kg$.
To find the ratio $m_e : m_p : m_n$,we divide each mass by the mass of the electron $(m_e)$:
Ratio $= 1 : \frac{1.672 \times 10^{-27}}{9.109 \times 10^{-31}} : \frac{1.674 \times 10^{-27}}{9.109 \times 10^{-31}}$
Ratio $\approx 1 : 1836.15 : 1838.68$.

(C) The correct matches are:
$A$. Chadwick discovered the neutron $(A-3)$.
$B$. Rutherford proposed the nuclear model of the atom $(B-4)$.
$C$. Moseley established the relationship between atomic number and $X$-ray spectra $(C-2)$.
$D$. $J$. $J$. Thomson performed the cathode ray experiment $(D-1)$.
Therefore,the correct sequence is $A-3, B-4, C-2, D-1$.

(C) The statement that cathode rays travel in a straight line in the presence of electric and magnetic fields is incorrect.
Cathode rays are deflected by both electric and magnetic fields.
When passed between two electrically charged plates,they are deflected towards the positively charged plate.
In a magnetic field,they are deflected at a right angle to both the field and the direction of the particle's motion.

(C) In Millikan's oil drop experiment,an electrically charged droplet of oil is placed between two parallel metal plates.
$1$. Gravitational force $(F_g = mg)$ acts downwards.
$2$. Viscous force $(F_v = 6 \pi \eta r v)$ acts opposite to the direction of motion of the droplet.
$3$. Electrostatic force $(F_e = qE)$ acts on the charged droplet when an electric field is applied.
Magnetic force is not involved in this experiment as there is no magnetic field present.
Therefore,the correct option is $(C)$.

(A) The $e/m$ ratio (specific charge) is calculated as the ratio of charge $(e)$ to mass $(m)$.
For a proton $(H^+)$,$e/m = 1/1 = 1$.
For a deuterium ion $(D^+)$,charge is $1$ and mass is $2$,so $e/m = 1/2 = 0.5$.
For an $\alpha$-particle $(He^{2+})$,charge is $2$ and mass is $4$,so $e/m = 2/4 = 0.5$.
Since both the deuterium ion and the $\alpha$-particle have an $e/m$ ratio of $0.5$,they have identical $e/m$ values.