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

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

  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.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

  1. $44$

  2. $54$

  3. $64$

  4. $74$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $54eV$.

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

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

  1. $40$

  2. $50$

  3. $60$

  4. $70$
Reveal answer Fill a bubble to check yourself
B Correct answer
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  $50^o$.

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

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

  1. $44, 68$

  2. $54, 78$

  3. $44, 58$

  4. $85, 100$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $54eV$.

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

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

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

  2. $5\sqrt{3}J$

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

  4. $3\sqrt {5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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

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

  1. $q.E$ and max

  2. $2q.E$ and min.

  3. $q.E$ and min

  4. zero and min.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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 } $

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

  2. zero

  3. $\pi$

  4. $\cfrac{3\pi}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\theta=\pi$

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

 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:

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

  2. $0$

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

  4. $\frac { p } { 8 \pi \epsilon _ { 0 } r ^ { 2 } }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The point (r/2, sqrt(3)r/2, 0) forms an equilateral triangle with the two dipoles at (0,0,0) and (r,0,0). The potential from the first dipole is V1 = (p cos theta1) / (4 pi epsilon0 r1^2) and from the second is V2 = (p cos theta2) / (4 pi epsilon0 r2^2). Summing these at the given coordinates yields the result.

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

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


  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)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work done in rotating a dipole in an electric field is W = U_final - U_initial. U = -pE cos(theta). Initial angle is 90 degrees (cos 90 = 0). Final angle is theta. W = -pE cos(theta) - 0 = -pE cos(theta).