2 Marks Questions

Question 12 Marks

Suppose a pure Si crystal has $5 \times 10^{28}$ atom $m ^{-3}$. It is doped by 1 ppm concentration of pentavalert As. Calculate the number of electrons and holes. (Given that $n_i=1.5 \times 10^{16} m^{-13}$ )

Answer

→ number of atoms in pure $S i=5 \times 10^{28} m^{-3}$
→ Concentration of $As =1 ppm$
Impurity proportion is kept as 1 in $10^6$ pure atoms.
$
\text { Total atoms of As }=\frac{5 \times 10^{28}}{10^6}=5 \times 10^{22} m^{-3}
$
→ As is pentavalent impurity. Thus As is donating one extra electron. So electron number density due to As atom,
$
n_e=n_{D}=5 \times 10^{22} m^{-3}
$
→ Now $n_i^2=n_e n_h$
$
\begin{array}{l}
\therefore n_h=\frac{n_i^2}{n_e} \\
\therefore n_h=\frac{\left(1.5 \times 10^{16}\right)^2}{5 \times 10^{22}} \\
=\frac{2.25 \times 10^{32}}{5 \times 10^{22}} \\
\therefore n_h \\
\therefore n_h=4.5 \times 10^9 m^{-3}
\end{array}
$

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Question 22 Marks

Write any two features of nuclear binding force.

Answer

→ The nucleus is made up of neutrons and protons. Therefore, it may be expected that the mass of the nucleus is equal to the total mass of its individual protons and neutrons.
→ But the nuclear mass $M$ is found to be always less than the total mass of its individual protons and neutrons.
→ For example :
${ }_8 O ^{16}$, a nucleus which has 8 neutrons and 8 protons.
Mass of 8 neutrons $=8 \cdot 1.00866 u$

Mass of 8 protons $=8 \cdot 1.00727 u$

Mass of 8 electrons $=8 \cdot 0.00055 u$
→ Therefore, the expected mass of ${ }_8 O ^{16}$ nucleus
$\begin{array}{l}=(8 \cdot 1.00866+8 \cdot 1.00727) \\ =8(1.00866+1.00727) \\ =8 \cdot 2.01593 u \\ =16.12744 u\end{array}$
→ The atomic mass of ${ }_8 O ^{16}$ found from mass spectroscopy experiments is seen to be $15.99493 u$.
→ Subtracting the mass of 8 electrons
$(8 \cdot 0.00055 u=0.0044 u)$ from this we get the experimental mass of ${ }_8 O ^{16}$ nucleus to be $15.99053 u$.
→ Thus, the mass of the ${ }_8 O ^{16}$ nucleus is less than the total mass of its constituents by
$
(16.12744-15.99053)=0.13691 u
$
→ "The difference in mass of a nucleus and its constituents, $\Delta M$ is called the mass defect" and is given by
$
\Delta M=\left[Z m_p+(A-Z) m_n\right]-M
$
Where, $Z =$ number of protons
$A - Z = N =$ neutron number
$m_p$ - mass of proton
$m_n$ - mass of neutron
M - mass of a nucleus
→ The energy equivalent to this mass defect is called the binding energy of nucleus.
$\therefore$ Binding energy $E _b=\Delta M c^2$
→ Binding energy per nucleon is the binding energy divided by the total number of nucleons.
$
\therefore E_{b n}=\frac{E_b}{A}
$
→ The binding energy per nucleon gives a measure of the stability of the nucleus.
→ A nucleus for which the value of $E _{b n}$ is comparatively higher is said to be more stable and for a nucleus for which the value of $E _{b n}$ is comparatively less is said to be less stable.

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Question 32 Marks

A hydrogen atom initially in the ground level absorbs a photon which excites it to the $n=4$ level. Determine the wavelength and frequency of photon.
\[\left(h=6.625 \times 10^{-34} Js, c=3 \times 10^8 m / s\right)\]

Answer

$
\begin{array}{l}
n=4 \\
\lambda=? \\
v=?
\end{array}
$
→ Total energy of electron
$
E_n=-\frac{13.6}{n^2} e V \ldots (1)
$
→ For ground state, $n=1$
$\begin{array}{l} E _1=-\frac{13.6}{1^2} eV \\ =-13.6 eV \end{array}$
$\begin{array}{l}\text { → Using } n=4 \text { in equation (1) } \\ \qquad \begin{aligned} E _4 & =-\frac{-13.6}{(4)^2} eV \\ = & \frac{-13.6}{16} eV \\ E _4 & =-0.85 eV \end{aligned}\end{array}$
→ Energy of incident photon
$\begin{array}{l} E _4- E _1=-0.85-(-13.6) \\ E _4- E _1=12.75 eV \\ \text { But, } E _i- E _f=h v \\ h v=12.75 eV \\ \quad \therefore v=\frac{12.75 \times 1.6 \times 10^{-19}}{6.625 \times 10^{-34}} \\ \therefore v=3.08 \times 10^{15} Hz\end{array}$
→ Wavelength of incident radiation
$\begin{array}{l}\lambda=\frac{c}{v}=\frac{3 \times 10^8}{3.08 \times 10^{15}} \\ \therefore \lambda=0.974 \times 10^{-7} m \\ \therefore \lambda=97.4 nm\end{array}$

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Question 42 Marks

Light of frequency $7.21 \times 10^{14} Hz$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^5 m / s$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?
\[\begin{array}{l}\left(h=6.625 \times 10^{-34} Js,\right. \\e=1.6 \times 10^{-19} C \\\left.m=9.1 \times 10^{-31} kg\right)\end{array}\]

Answer

$
\begin{array}{l}
→ v=7.21 \times 10^{14} Hz \\
v_{\max }=6.0 \times 10^5 m / s \\
v_0=? \\
K_{\max }=h v-\varphi_0
\end{array}
$
$
\begin{array}{l}
\therefore \frac{1}{2} m v_{\max }^2=h v-\varphi_0\left({ }^{\prime} K_{\max }=\frac{1}{2} m v_{\max }^2\right) \\
\therefore \varphi_0=h v-\frac{1}{2} m v_{\max }^2 \\
\therefore h v_0=h v-\frac{1}{2} m v_{\max }^2\left({ }^{\prime} \varphi_0=h v_0\right) \\
\therefore v_0=v-\frac{mv_{\max }^2}{2 h} \\
\therefore v=\left(7.21 \times 10^{14}\right)-\left(\frac{9.1 \times 10^{-31} \times\left(6.0 \times 10^5\right)^2}{2 \times 6.625 \times 10^{-34}}\right) \\
v_0=\left(7.21 \times 10^{14}\right)-\left(2.472 \times 10^{14}\right) \\
v_0=4.738 \times 10^{14} Hz
\end{array}
$

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Question 52 Marks

Using Huygens principle, explain reflection of a plane wave by a plane reflecting surface.

Answer

→ Consider a plane wave AB incident at an angle $i$ on a reflecting surface MN .
→ If $u$ is the speed of wave in the given medium, $\tau$ represents the time taken by the wavefront to advance from point $B$ to $C$, then the distance $BC =v \tau$.
→ In order to construct the reflected wavefront we draw a sphere of radius $v$ from the point $A$ as shown in fig.
→ Let CE represent the tangent plane drawn from the point C to this sphere.
→ Obviously,
$
AE=BC=UT
$
→ From fig., incident and reflected wave fronts make angle $i$ and $r$ with reflecting surface, MN respectively.
→ From fig.,
AC is the common side between $\triangle AEC$ and $\triangle ABC$,
Also, $\angle AEC =\angle ABC =\frac{\pi}{2}$
$
\text { and } AE=BC=UT
$
→ So $\triangle AEC$ and $\triangle ABC$ are congruent.
→ And therfore, $\angle i=\angle r$,
which is law of reflection.

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Question 62 Marks

Derive the relation between focal length $(f)$ and radius of curvature ( R ) for a spherical convex mirror with the help of a geometrical diagram of reflection of incident ray on a convex spherical mirror.

Answer

→ As shown in figure, C is center of curvature of the mirror and F is the principal focus of the mirror.
→ As shown in figure, a ray is incident at an angle $\theta$ near a point M on the mirror. CM is the perpendicular drawn to the surface. $M D$ is the perpendicular drawn from point $M$ on the principal axis.
→ From figure, $\angle MCP =\theta$ and $\angle MFP =2 \theta$
→ Now, $\tan \theta=\frac{M D}{C D} \ldots(1)$
and $\tan 2 \theta=\frac{ MD }{ FD } \ldots(2)$
→ Since $\theta$ is extremely small for paraxial rays
$
\tan \theta \approx \theta \text { and } \tan 2 \theta \approx 2 \theta
$
→ From equation (1) and equation (2),
$
\begin{array}{l}
\theta=\frac{MD}{CD} \text { and } 2 \theta=\frac{MD}{FD} \\
\therefore 2\left(\frac{MD}{CD}\right)=\frac{MD}{FD} \\
\therefore \frac{2}{CD}=\frac{1}{FD} \\
\therefore C D=2 FD \ldots(3)
\end{array}
$
→ Now for small $\theta$ the point $D$ is very close to the point $P$.
→ Therefore, $FD = FP =f$ and $CD = CP = R$
Substituting above values in equation (3),
$
\begin{array}{l}
R=2 f \\
\text { or } f=\frac{R}{2}
\end{array}
$

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Question 72 Marks

A $100 \Omega$ resistor is connected to a $220 V, 50 Hz$ AC supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?

Answer

$
\begin{array}{l}
V= 220 V \\
v=50 Hz \\
R=100 \Omega
\end{array}
$
(a) rms value of current in circuit,
$
\begin{array}{l}
I=\frac{V}{R} \\
\therefore I=\frac{220}{100} \\
\therefore I=2.2 A
\end{array}
$
(b) Power consumed over a full cycle,
$
\begin{array}{l}
P=VI \\
\therefore P=(220)(2.2) \\
\therefore P=484 W
\end{array}
$

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Question 82 Marks

Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B , area A and length $l$ of the solenoid and hence obtain formula for magnetic energy per unit volume.

Answer

→ (a) Magnetic energy stored in solenoid
$
\begin{array}{l}
U_{B}=\frac{1}{2} LI^2 \\
=\frac{1}{2} L^{\left(\frac{B}{\mu_0 n}\right)^2}
\end{array}
$
( $\because$ for solenoid, $B =\propto_0 n I \Rightarrow I =\frac{ B }{\mu_0 n}$ )
$
=\frac{1}{2}\left(\propto_0 n^2 A l\right)\left(\frac{B}{\mu_0 n}\right)^2
$
( $\because$ Self inductance of solenoid $L =\propto_0 n^2 A l$ )
$
\begin{array}{l}
=\frac{1}{2} \propto_0 n^2 A l \times \frac{B^2}{\mu_0^2 n^2} \\
U_{B}=\frac{1}{2 \mu_0}\left(B^2 A l\right) \ldots(1)
\end{array}
$
(b) Magnetic energy per unit volume
$
\begin{array}{l}
\rho_{B}=\frac{U_{B}}{V}(\because V=\text { Volume }) \\
=\frac{U_{B}}{A l}(\because \text { Volume } V=A l) \\
=\frac{1}{2 \mu_0} \frac{B^2 A l}{Al} \text { (from eq. 1) } \\
\rho_{B}=\frac{B^2}{2 \mu_0} \ldots \text { (2) }
\end{array}
$
→ Electrostatic energy stored per unit volume in parallel plate capacitor.
$
\rho_{E}=\frac{1}{2} \varepsilon_0 E^2 \ldots(3)
$
→ From equation (2) \& (3), energy is directly proportional to square of field intensity in both cases.

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Question 92 Marks

Write the characteristics of magnetic field lines.

Answer

→ Characteristics of magnetic field lines are as follows:
(i) The magnetic field lines form continuous closed loops. In the region outside a magnet, the magnetic field lines are from N to S , whereas in the region inside the magnet, the field lines are from S to N .
(ii) The tangent to the field line at a given point represents the direction of the magnetic field ( $\overrightarrow{ B }$ ) at that point.
(iii) The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field $\vec{B}$.
(iv) Two magnetic field lines do not intersect each other. Because if they do, at the point of intersection, the magnetic field would have two directions, which is not possible.

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Question 102 Marks

Define mobility and write its formula, unit and dimensional formula.

Answer

→ Conductivity of conductor arises from mobile charge carriers.
→ In metals, electrons are mobile charge carriers. In an electrolyte they are positive and negative ions while in an ionised gas, they are electrons and positive charged ions. In semi conductors, free electrons and holes are mobile charge carriers.
→ "The magnitude of the drift velocity per unit electric field is called the mobility ( $\infty$ )."
$
\infty=\frac{\left|\overrightarrow{v_d}\right|}{E} \ldots
$
→ The SI unit of mobility is $m^2 / Vs$ and practical unit is $cm ^2 / Vs$
→ Dimensional formula of mobility is $M ^{-1} L^0 T^2 A^1$
→ Using drift velocity $\left|\vec{v}_d\right|=\frac{ Ee }{m} \cdot \tau$
in equation (1).
we get,
$
\begin{array}{l}
\propto=\frac{\frac{e E}{m} \tau}{E} \\
\therefore \propto=\frac{e \tau}{m}
\end{array}
$
where $\tau$ is the average collision time for electrons.

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Question 112 Marks

An infinite line charge produces a field of $9 \times 10^4$ $N / C$ at a distance of 2 cm . Calculate the linear charge density.

Answer

→ $
r=2 cm=2 \cdot 10^{-2} m
$
$
\begin{array}{l}
E=9 \cdot 10^4 N / C \\
\lambda=?
\end{array}
$
→ As per the formula, electric field at a distance ' $r$ ' from a very long (infinite) line charge,
$
E=\frac{2 k \lambda}{r}
$
$\begin{array}{l}\therefore \lambda=\frac{ Er }{2 k} \\ =\frac{9 \times 10^4 \times 2 \times 10^{-2}}{2 \times 9 \times 10^9} \\ \therefore \lambda=0.1 \cdot 10^{-6} \\ \therefore \lambda=0.1 \propto C / m \end{array}$

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Question 122 Marks

Derive the formula for the torque acting on a dipole placed in a uniform external electric field.

Answer

→ As shown in figure, the electric dipole is placed in uniform electric field at $\theta$ angle.
→The force exerted on $+q$ electric charge in electric field $\overrightarrow{ E }_{\text {is }}$,
$
\overrightarrow{F}+=q \overrightarrow{E}
$
→ The force exerted on - $q$ electric charge
$
\overrightarrow{F}_{-}=-q \overrightarrow{E}
$
→ The net force on the dipole is zero, since $E$ is uniform.
→ However, the charges are separated, so the forces act at different points, resulting in a torque on the dipole.
→ When the net force is zero, the torque (couple) is independent of the origin. Its magnitude equals the magnitude of each force multiplied by the arm of the couple (perpendicular distance between the two antiparallel forces).
→ Magnitude of torque $=q E \times 2 a \sin \theta$
$
\begin{array}{l}
=(2 q a) E \sin \theta=p E \sin \theta \\
\vec{\tau}=\vec{p} \cdot \overrightarrow{E}
\end{array}
$
→ The magnitude of $\vec{p} \times \overrightarrow{ E }$ is also $p E \sin \theta$ and its direction is normal to the paper, coming out of it.
→ Special cases :
(i) If electric dipole moment and electric field both are in one direction.
$
\begin{array}{l}
\therefore \theta=0 \\
\therefore \tau=0
\end{array}
$
This condition is called stable equilibrium.
(ii) Both electric dipole moment and electric field are perpendicular.
$
\therefore \theta=\frac{\pi}{2}
$
$\therefore \tau=p E \sin \frac{\pi}{2}$
$\therefore \tau=p E \rightarrow$ Which is maximum
(iii) Electric dipole moment and electric field are arranged anti parallel direction.
$
\begin{array}{l}
\therefore \tau=pE \sin \pi \\
\therefore \tau=0 \rightarrow \text { Unstable equilibrium }
\end{array}
$

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Physics 2 Marks Questions

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