Tag: physics

Questions Related to physics

Electrons can be diffracted (Davis-son and German's expt.).

physics dual nature of matter and radiation davisson and germer experiment and its conclusion matter waves wave nature of matter

  1. Yes, as their wave is transverse.

  2. Yes, as their wave is longitudinal

  3. No, as their wave is longitudinal

  4. No, as they travel in a straight line.

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Correct Option: D
Explanation:

As electron waves oscillation is confined in one orientation only, therefore  it cannot be polarized.

In Davisson-Germer experiment, intensity was maximum for accelerating voltage equal to

physics dual nature of matter and radiation davisson and germer experiment and its conclusion matter waves wave nature of matter

  1. $44$

  2. $54$

  3. $64$

  4. $74$

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Correct Option: B
Explanation:

In Davisson-Germer experiment, maximum intensity of diffracted electron beam was found at different angles by varying the applied voltage to the electron gun. The highest intensity was observed at an angle $\phi =50^o$ with a voltage of $54 V$, giving the electron a kinetic energy of $54eV$.

In Davisson-Germer experiment, intensity was maximum for scattering angle equal to

physics dual nature of matter and radiation davisson and germer experiment and its conclusion matter waves wave nature of matter

  1. $40$

  2. $50$

  3. $60$

  4. $70$

Show answer
Correct Option: B
Explanation:

In Davisson -Germer experiment, it was observed that the intensity of the scattered electron beam depends on the scattering angle $\phi$. Also, it Davisson and Germer observed that the maximum intensity was detected when the scattering angle was  $50^o$.

The Davisson-Germer experiment was performed by varying the accelarating voltage from __ V to __ V.

physics dual nature of matter and radiation davisson and germer experiment and its conclusion matter waves wave nature of matter

  1. $44, 68$

  2. $54, 78$

  3. $44, 58$

  4. $85, 100$

Show answer
Correct Option: A
Explanation:

In Davisson-Germer experiment, maximum intensity of diffracted electron beam was found at different angles by varying the applied voltage to the electron gun from $44V$ to $68V$. The highest intensity was observed at an angle $\phi =50^o$ with a voltage of $54 V$, giving the electron a kinetic energy of $54eV$.

An electric dipole of length $20cm$ having $\pm 3\times { 10 }^{ -3 }C$ charge placed at ${60}^{o}$ with respect to a uniform electric field experiences a torque of magnitude $6Nm$. The potential energy of the dipole is

physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

  1. $-2\sqrt{3}J$

  2. $5\sqrt{3}J$

  3. $-2\sqrt {2}J$

  4. $3\sqrt {5}$

Show answer
Correct Option: A
Explanation:

Here length of dipole $2a=20cm=20\times { 10 }^{ -2 }m$, Charge $q=\pm 3\times { 10 }^{ -3 }C,\theta ={ 60 }^{ o }\quad $ and torque $\tau =6Nm$
As $\tau =pE\sin { \theta  } $
or $E=\cfrac { \tau  }{ p\sin { \theta  }  } =\cfrac { \tau  }{ q(2a)\sin { \theta  }  } \left( \because p=q(2a) \right) $
$\therefore E=\cfrac { 6 }{ 3\times { 10 }^{ -3 }\times 20\times { 10 }^{ -2 }\times \sin { { 60 }^{ o } }  } =\cfrac { { 10 }^{ 5 } }{ 5\sqrt { 3 }  } N{ C }^{ -1 }$
Potential energy of dipole $U=-pE\cos{\theta}=-q(2a)E\cos{\theta}$
$=-3\times { 10 }^{ -3 }\left( 20\times { 10 }^{ -2 } \right) \cfrac { { 10 }^{ 5 } }{ 5\sqrt { 3 } } \cos { { 60 }^{ o } } =\cfrac { -3\times { 10 }^{ -5 }\times 20\times { 10 }^{ 5 } }{ 5\sqrt { 3 } \times 2 } =-2\sqrt { 3 } J\quad \quad $

An electric dipole has the magnitude of its charge as $q$ and its dipole moment is $p$. It is placed in uniform electric field $E$. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively

physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

  1. $q.E$ and max

  2. $2q.E$ and min.

  3. $q.E$ and min

  4. zero and min.

Show answer
Correct Option: D
Explanation:

When the dipole is in the direction of field then net force is $qE+(-qE)=0$
and its potential energy is minimum $=-p.E$
$=-qaE$

An electric dipole of diploe moment $\overrightarrow { p } $ placed in uniform electric field $\overrightarrow { E } $ has minimum potential energy when angle between $\overrightarrow { p } $ and $\overrightarrow { E } $

physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

  1. $\cfrac{\pi}{2}$

  2. zero

  3. $\pi$

  4. $\cfrac{3\pi}{2}$

Show answer
Correct Option: C
Explanation:

Potential Energy=$ -PE \cos {\theta}$

when 
$ \theta=0 $
Potential Energy=$ -PE $
When
$ \theta=180 $
Potential Energy=$ +PE $
So, Maximum Potential Energy=$ +PE $ at angle $\theta=\pi$

 Two small electric dipoles each of dipole moment pi are situated at $(0, 0, 0)$ and $(r, 0, 0)$. the electric potential at a point $\left( \frac { r } { 2 } , \frac { \sqrt { 3 } r } { 2 } , 0 \right)$ is:

physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

  1. $\frac { p } { 4 \pi \in _ { 0 } r ^ { 2 } }$

  2. $0$

  3. $\frac { p } { 2 \pi \epsilon _ { 0 } r ^ { 2 } }$

  4. $\frac { p } { 8 \pi \epsilon _ { 0 } r ^ { 2 } }$

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Correct Option: D

Potential at any point in the electric field produced by a dipole is

physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

  1. $\infty , r$

  2. $\alpha r ^ { 2 }$

  3. $\frac { 1 } { r }$

  4. $\frac { 1 } { r ^ { 2 } }$

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Correct Option: A

A dipole of dipole moment $\overline {\text{p}} $ i s aligned at right angle to electrictric field $\overline {\text{E}} $ . To set it at an angle $\theta $ with E the amount of work done is


physics electric charges and fields potential energy of a dipole in external field potential due to electric dipole electric dipole

  1. $ - {\text{pEcos}}\theta $

  2. $ {\text{pEsin}}\theta $

  3. $ - {\text{pE}}\left( {{\text{sin}}\theta - 1} \right)$

  4. $ - {\text{pE}}\left( {{\text{sin}}\theta + 1} \right)$

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Correct Option: A