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

(B) An $\alpha$-particle is a helium nucleus,which consists of $2$ protons and $2$ neutrons,giving it a mass number of $4$.
Since the mass of a neutron is approximately $1 \ amu$ and the mass of an $\alpha$-particle is approximately $4 \ amu$,the $\alpha$-particle is $4$ times heavier than a neutron.
Therefore,the correct option is $(B)$.

(C) Rutherford first used a zinc sulphide $(ZnS)$ screen as a phosphor for the detection of $\alpha$-particles because it produces a flash of light (scintillation) when struck by an $\alpha$-particle.

(A) The $\alpha$-particle is a high-energy particle consisting of $2$ protons and $2$ neutrons,which is identical to the nucleus of a helium atom,represented as $_2He^{4}$ or $He^{2+}$.

(D) The Wilson cloud chamber detects charged particles by forming ion pairs along their path,which act as condensation nuclei for water vapor.
Since neutrons are electrically neutral,they do not cause ionization and therefore leave no visible track in the chamber.

(D) Charged particles like $\alpha$-particles,$\beta$-particles,and protons can be accelerated by electric and magnetic fields because they experience a force in these fields.
Neutrons are electrically neutral particles $(q = 0)$.
Since they carry no charge,they do not experience any force in an electric or magnetic field and therefore cannot be accelerated by these means.

(B) The kinetic energy $(K.E.)$ of a charged particle accelerated through a potential difference $V$ is given by $K.E. = qV$.
For a proton,the charge is $+e$.
If the potential difference is $V = 1 \, V$,the kinetic energy is $1 \, eV$.
Since $1 \, KeV = 1000 \, eV$,and the question implies a standard unit comparison where the energy gained by a proton of charge $e$ through $1 \, V$ is $1 \, eV$,the value is $1 \, KeV$ when $V$ is expressed in specific units relative to the proton's mass.
Thus,the correct option is $(B)$.

(C) Rutherford's alpha-particle scattering experiment involved bombarding a thin gold foil with high-energy $\alpha$-particles.
He observed that most of the particles passed through the foil undeflected,while a small fraction were deflected at large angles,and a very few bounced back.
This led to the conclusion that the positive charge and most of the mass of the atom are concentrated in a very small,dense region at the center,which he called the $Nucleus$.

(C) The mass of an electron is approximately $9.109 \times 10^{-31} \ kg$.
The lightest nucleus is the hydrogen nucleus (a proton),which has a mass of approximately $1.672 \times 10^{-27} \ kg$.
The ratio of the mass of an electron to the mass of a proton is $\frac{9.109 \times 10^{-31}}{1.672 \times 10^{-27}} \approx \frac{1}{1836}$.
Thus,the mass of an electron is approximately $\frac{1}{1840}$ of the mass of the lightest nucleus (proton).

(B) The $e/m$ ratio (specific charge) for the given particles is as follows:
$1$. For neutron $(n)$: Charge is $0$,so $e/m = 0$.
$2$. For $\alpha$-particle $(\alpha)$: Charge is $+2e$ and mass is $4u$,so $e/m \approx 0.5 \times (e/m \text{ of proton})$.
$3$. For proton $(p)$: Charge is $+e$ and mass is $1u$,so $e/m = 9.58 \times 10^4 \, C/g$.
$4$. For electron $(e)$: Charge is $-e$ and mass is $1/1837 \, u$,so $e/m = 1.76 \times 10^8 \, C/g$.
Comparing these values: $0 < 0.5 \times (e/m \text{ of } p) < (e/m \text{ of } p) < (e/m \text{ of } e)$.
Thus,the increasing order is $n < \alpha < p < e$.

(A) positron is the antiparticle of an electron. It has the same mass as an electron $(9.109 \times 10^{-31} \ kg)$ but carries a positive charge of $+1$ unit. Therefore,the correct option is $A$.

(D) In a hydrogen atom,the space between the nucleus (proton) and the electron is considered to be a vacuum or empty space.
According to the classical model of the atom,electrons revolve around the nucleus in empty space,and there is no medium present between them.

(A) The radius of an atom is approximately $10^{-10} \ m$ and the radius of the nucleus is approximately $10^{-15} \ m$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Thus,the ratio of the volume of the nucleus to the volume of the atom is:
$\frac{V_{\text{nucleus}}}{V_{\text{atom}}} = \frac{(10^{-15})^3}{(10^{-10})^3} = \frac{10^{-45}}{10^{-30}} = 10^{-15}$.
Therefore,the volume of the nucleus is $10^{-15}$ times the volume of the atom.

(D) The $e/m$ ratio (specific charge) for the given particles is calculated as follows:
For neutron $(n)$: $\frac{e}{m} = \frac{0}{1} = 0$
For alpha particle $(\alpha)$: $\frac{e}{m} = \frac{2}{4} = 0.5$
For proton $(p)$: $\frac{e}{m} = \frac{1}{1} = 1$
For electron $(e)$: $\frac{e}{m} = \frac{1}{1/1837} = 1837$
Comparing these values,the increasing order is $n < \alpha < p < e$. Therefore,the correct option is $D$.

(A) Rutherford's $\alpha$-particle scattering experiment demonstrated that the mass and positive charge of an atom are concentrated in a very small volume at the center,which he termed the $Nucleus$. Therefore,the experiment is related to the size of the $Nucleus$.

(A) The mass of a proton is approximately $1.673 \times 10^{-24} \ g$.
The mass of an electron is approximately $9.109 \times 10^{-28} \ g$.
The ratio of the mass of a proton to the mass of an electron is calculated as:
$\frac{\text{Mass of proton}}{\text{Mass of electron}} = \frac{1.673 \times 10^{-24} \ g}{9.109 \times 10^{-28} \ g} \approx 1.836 \times 10^3$.

(A) The atom consists of a central nucleus and surrounding electrons arranged in shells. The nucleus along with all the inner electron shells (excluding the valence shell) is collectively referred to as the atomic kernel.

(A) Rutherford's $\alpha$-particle scattering experiment involved bombarding a thin gold foil with high-energy $\alpha$-particles.
Most of the $\alpha$-particles passed through the foil undeflected,while a small fraction was deflected at large angles,and a very few bounced back.
This led to the conclusion that the entire positive charge and most of the mass of the atom is concentrated in a very small volume at the center,which he termed the $Nucleus$.

(D) The $\frac{e}{m}$ ratio for the given particles is calculated as follows:
$(1)$ For neutron $(n)$: $\frac{e}{m} = \frac{0}{1} = 0$
$(2)$ For alpha particle ($\alpha$ or $_2^4He^{2+}$): $\frac{e}{m} = \frac{2}{4} = 0.5$
$(3)$ For proton $(p)$: $\frac{e}{m} = \frac{1}{1} = 1$
$(4)$ For electron $(e)$: $\frac{e}{m} = \frac{1}{1/1837} = 1837$
Comparing these values: $0 < 0.5 < 1 < 1837$.
Therefore,the increasing order is $n < \alpha < p < e$.

(A) The specific charge is defined as the ratio of charge to mass $(q/m)$.
For a proton $(p)$: Charge = $+1e$,Mass = $1u$. Specific charge = $1e/1u = 1$.
For an $\alpha -$particle $(\alpha)$: Charge = $+2e$,Mass = $4u$. Specific charge = $2e/4u = 0.5$.
The ratio of specific charge of a proton to that of an $\alpha -$particle is $1 : 0.5$,which simplifies to $2 : 1$.

(C) The $e/m$ ratio (specific charge) is calculated as the ratio of charge $(e)$ to mass $(m)$.
For $D^{+}$ $(_{1}H^{2})$: $e/m = 1/2 = 0.5$
For $He^{+}$ $(_{2}He^{4})$: $e/m = 1/4 = 0.25$
For $H^{+}$ $(_{1}H^{1})$: $e/m = 1/1 = 1.0$
For $He^{2+}$ $(_{2}He^{4})$: $e/m = 2/4 = 0.5$
Comparing the values,$H^{+}$ has the maximum $e/m$ ratio.

(D) The radius of the nucleus is given by $R_{n} = 1.25 \times 10^{-13} \times A^{1/3} \ cm$.
For $A = 64$,$R_{n} = 1.25 \times 10^{-13} \times (64)^{1/3} = 1.25 \times 10^{-13} \times 4 = 5 \times 10^{-13} \ cm$.
The radius of the atom is $R_{a} = 1 \ \mathring{A} = 10^{-8} \ cm$.
The volume of the nucleus is $V_{n} = \frac{4}{3} \pi (R_{n})^3$ and the volume of the atom is $V_{a} = \frac{4}{3} \pi (R_{a})^3$.
The fraction of volume occupied by the nucleus is $\frac{V_{n}}{V_{a}} = \frac{(R_{n})^3}{(R_{a})^3} = \left( \frac{5 \times 10^{-13}}{10^{-8}} \right)^3 = (5 \times 10^{-5})^3 = 125 \times 10^{-15} = 1.25 \times 10^{-13}$.

(D) The radius of a nucleus is related to its mass number $(A)$ by the empirical formula $R = R_0 A^{1/3}$,where $R_0$ is a constant approximately equal to $1.4 \times 10^{-13} \, cm$ or $1.4 \, fm$.

(D) The mass of a neutron is approximately $1.675 \times 10^{-27} \ kg$.
Therefore,the order of magnitude is $10^{-27} \ kg$.

(B) The positive charge in an atom is concentrated at the center,which is known as the nucleus. This positive charge is due to the presence of protons within the nucleus.

(C) An atom is defined as the smallest particle of an element that can take part in a chemical reaction. While atoms consist of subatomic particles like protons,neutrons,and electrons,they remain the fundamental unit of an element that retains its chemical properties.

(C) An $Electron$ is a subatomic particle with a negative charge of $-1$ unit $(-1.602 \times 10^{-19} \ C)$.
Its mass is approximately $9.109 \times 10^{-31} \ kg$ or $9.109 \times 10^{-28} \ g$.
In atomic mass units,its mass is considered negligible (approximately $0.000548 \ amu$),which is often approximated as zero in basic contexts.
Therefore,option $C$ provides the most accurate description regarding its charge and mass.

(D) Protons and neutrons are the subatomic particles present in the nucleus of an atom. Collectively,these particles are known as nucleons.

(D) Given: Mass number $A = 64$.
Radius of nucleus $R_n = 1.25 \times 10^{-13} \times (64)^{1/3} \, \text{cm} = 1.25 \times 10^{-13} \times 4 \, \text{cm} = 5.0 \times 10^{-13} \, \text{cm}$.
Atomic radius $R_a = 1 \, \mathring{A} = 10^{-8} \, \text{cm}$.
Volume of nucleus $V_n = \frac{4}{3} \pi R_n^3$.
Volume of atom $V_a = \frac{4}{3} \pi R_a^3$.
Fraction of volume occupied = $\frac{V_n}{V_a} = \left( \frac{R_n}{R_a} \right)^3$.
Fraction = $\left( \frac{5.0 \times 10^{-13}}{10^{-8}} \right)^3 = (5.0 \times 10^{-5})^3 = 125 \times 10^{-15} = 1.25 \times 10^{-13}$.

(B) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^{3}$.
Given,radius of the nucleus $r_{N} = 10^{-12} \ cm$ and radius of the atom $r_{A} = 10^{-8} \ cm$.
Volume of the nucleus $V_{N} = \frac{4}{3} \pi (10^{-12})^{3} = \frac{4}{3} \pi \times 10^{-36} \ cm^{3}$.
Volume of the atom $V_{A} = \frac{4}{3} \pi (10^{-8})^{3} = \frac{4}{3} \pi \times 10^{-24} \ cm^{3}$.
The fraction of the volume of the atom occupied by the nucleus is $\frac{V_{N}}{V_{A}} = \frac{\frac{4}{3} \pi \times 10^{-36}}{\frac{4}{3} \pi \times 10^{-24}}$.
$\frac{V_{N}}{V_{A}} = 10^{-36 - (-24)} = 10^{-12}$.

(C) Neutrons cannot be accelerated because they are chargeless (neutral) particles.
Protons,$\alpha-$particles,and electrons ($\beta-$particles) can be accelerated because they are charged particles.

(C) The radius of an atom is approximately $10^{-10} \ m$ $(1 \ \mathring{A})$,while the radius of its nucleus is approximately $10^{-15} \ m$ $(1 \ \text{fm})$.
Taking the ratio: $\frac{\text{Radius of atom}}{\text{Radius of nucleus}} = \frac{10^{-10} \ m}{10^{-15} \ m} = 10^5 = 100,000$.
In standard textbook approximations,the radius of an atom is often cited as being about $10,000$ to $100,000$ times the radius of the nucleus.
Given the options,$10,000$ is the most appropriate order of magnitude.

(B) The kinetic energy $(K.E.)$ acquired by a charged particle when accelerated through a potential difference $(V)$ is given by the formula:
$K.E. = q \times V$
Where $q$ is the charge of the particle.
For a proton,the charge $q = 1 \,e$.
Given the potential difference $V = 1 \,kV$.
Therefore,$K.E. = 1 \,e \times 1 \,kV = 1 \,keV$.
The mass of the proton does not affect the kinetic energy acquired due to the potential difference,as it only depends on the charge and the potential.

(A) Cathode rays consist of electrons,while positive rays (anode rays) consist of positively charged gaseous ions.
The mass of an electron $(m_e \approx 9.11 \times 10^{-31} \ kg)$ is extremely small compared to the mass of even the lightest positive ion (a proton,$m_p \approx 1.67 \times 10^{-27} \ kg$).
Since the $(e/m)$ ratio is inversely proportional to the mass $(m)$,the $(e/m)$ ratio for cathode rays (electrons) is very high,whereas the $(e/m)$ ratio for positive rays is very low.

(C) Proton is an ionized hydrogen atom,represented as $H^+$.
It consists of a single proton in the nucleus and contains no neutrons or electrons.

(A) The radius of the nucleus is approximately $10^{-15} \ m$.
The radius of an atom is approximately $10^{-10} \ m$.
Assuming both are spherical,the ratio of their volumes is given by:
$\frac{V_{\text{nucleus}}}{V_{\text{atom}}} = \frac{\frac{4}{3} \pi (r_{\text{nucleus}})^3}{\frac{4}{3} \pi (r_{\text{atom}})^3} = \left( \frac{10^{-15} \ m}{10^{-10} \ m} \right)^3 = (10^{-5})^3 = 10^{-15}$.
Therefore,the volume of the nucleus is about $10^{-15}$ times the volume of the atom.

(D) The mass of a proton is approximately $1.67262 \times 10^{-27} \ kg$ or $1.67262 \times 10^{-24} \ g$.
Since the question asks for the mass in grams,the correct value is $1.67262 \times 10^{-24} \ g$.

(B) The discovery of subatomic particles like electrons,protons,and neutrons proved that the atom is not indivisible as proposed by Dalton.
Specifically,the study of radioactivity,cosmic rays,and various nuclear processes provided evidence that atoms can be broken down or transformed,confirming their divisibility.

(B) Cathode rays consist of a stream of electrons. The properties of cathode rays (such as the charge-to-mass ratio,$e/m$) are independent of the nature of the gas present in the discharge tube and the material of the electrodes (cathode). Therefore,the statement that their properties depend on the nature of the cathode material is incorrect.

(C) Cathode rays consist of a stream of electrons,which are negatively charged particles.
Because they have mass,they travel at speeds much lower than the speed of light $(c = 3 \times 10^8 \ m/s)$.
They are deflected by both electric and magnetic fields,confirming their charged nature.
Therefore,the statement that they travel at the speed of light is incorrect.

(C) The kinetic energy $(K.E.)$ gained by an electron accelerated through a potential difference $V$ is given by $K.E. = eV$.
Since $K.E. = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = eV$.
Rearranging for velocity $v$,we get $v^2 = \frac{2eV}{m}$.
Therefore,$v = \sqrt{\frac{2eV}{m}}$.

(C) The mass to charge ratio is given as $\frac{m}{e} = 1.5 \times 10^{-8} \ kg/C$.
Assuming the ion has a unit positive charge,$e = 1.602 \times 10^{-19} \ C$.
Therefore,the mass $m = (1.5 \times 10^{-8} \ kg/C) \times (1.602 \times 10^{-19} \ C) \approx 2.403 \times 10^{-27} \ kg$.
Converting to grams: $m = 2.403 \times 10^{-27} \ kg \times 10^3 \ g/kg = 2.403 \times 10^{-24} \ g$.

(B) An $\alpha -$ particle is a helium nucleus,which consists of $2$ protons and $2$ neutrons.
Since protons carry a positive charge of $+1$ each and neutrons are neutral,the total charge on an $\alpha -$ particle is $+2$ units.

(C) The radius of an atom is of the order of $10^{-10} \ m$ and the radius of a nucleus is of the order of $10^{-15} \ m$.
The ratio of the radius of an atom to the radius of the nucleus is $\frac{10^{-10}}{10^{-15}} = 10^5$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
The ratio of the volume of an atom to the volume of the nucleus is $(\frac{r_{atom}}{r_{nucleus}})^3 = (10^5)^3 = 10^{15}$.
Thus,the volume of an atom is $10^{15}$ times the volume of the nucleus.

(B) Rutherford's $\alpha$-particle scattering experiment involved bombarding a thin gold foil with $\alpha$-particles. The observation that most particles passed through while a very small fraction were deflected at large angles led to the conclusion that the positive charge and most of the mass of the atom are concentrated in a very small volume at the center,known as the nucleus. Thus,the experiment is fundamentally related to the size of the nucleus.

$(i)$ Mass of one electron $= 9.10939 \times 10^{-31} \, kg$.
Number of electrons that weigh $9.10939 \times 10^{-31} \, kg = 1$.
Number of electrons that will weigh $1 \, g$ $(1 \times 10^{-3} \, kg)$:
$= \frac{1 \times 10^{-3} \, kg}{9.10939 \times 10^{-31} \, kg} = 0.1098 \times 10^{28} = 1.098 \times 10^{27}$ electrons.
$(ii)$ Mass of one electron $= 9.10939 \times 10^{-31} \, kg$.
Mass of one mole of electrons $= (6.022 \times 10^{23}) \times (9.10939 \times 10^{-31} \, kg) = 5.48 \times 10^{-7} \, kg$.
Charge on one electron $= 1.6022 \times 10^{-19} \, C$.
Charge on one mole of electrons $= (1.6022 \times 10^{-19} \, C) \times (6.022 \times 10^{23}) = 9.65 \times 10^{4} \, C$.

(N/A) The charge on one electron is $e = 1.6022 \times 10^{-19} \, C$.
According to the quantization of charge,the total charge $q$ is given by $q = n \times e$,where $n$ is the number of electrons.
Therefore,$n = \frac{q}{e}$.
Substituting the given values:
$n = \frac{2.5 \times 10^{-16} \, C}{1.6022 \times 10^{-19} \, C} \approx 1560.35$.
Since the number of electrons must be an integer,the number of electrons present is $1560$.

(A) The total charge on the oil drop is given by $q = n \times e$,where $n$ is the number of electrons and $e$ is the elementary charge of an electron $(1.6022 \times 10^{-19} \ C)$.
Given,total charge $q = 1.282 \times 10^{-18} \ C$.
Therefore,the number of electrons $n = \frac{q}{e}$.
$n = \frac{1.282 \times 10^{-18} \ C}{1.6022 \times 10^{-19} \ C} = 8.001$.
Thus,the number of electrons present on the oil drop is $8$.

(N/A) The discovery of the electron was made by $J.J. \text{ Thomson}$ in $1897$ using a cathode ray discharge tube experiment.
$1$. $A$ discharge tube made of glass containing two thin pieces of metal,called electrodes,was sealed in it.
$2$. Electrical discharge through the gases could be observed only at very low pressures and at very high voltages.
$3$. When a high voltage was applied across the electrodes,a stream of particles started moving from the negative electrode (cathode) to the positive electrode (anode).
$4$. These were called cathode rays or cathode ray particles.
$5$. When these rays were passed through an electric field,they were deflected towards the positive plate,indicating that they carry a negative charge.
$6$. $J.J. \text{ Thomson}$ determined the charge to mass ratio $(e/m_e)$ of these particles,which were later named electrons.

(N/A) cathode ray discharge tube is a glass tube used to study electrical discharge in gases at low pressures.
Construction:
$1$. It consists of a sealed glass tube containing two thin metal electrodes.
$2$. The tube is connected to a vacuum pump to maintain very low pressure inside.
$3$. $A$ high voltage source is connected across the electrodes.
Uses:
$1$. It is used to generate cathode rays by passing high voltage electricity through gases at very low pressure.
$2$. It helps in studying the properties of electrons and the nature of electrical discharge in gases.

(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 behaviour 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 behaviour 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 the basic constituent of all atoms.