Dual Nature of Radiation and Matter — Page 7 | NEET Physics

Q61

A beam of electron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move

A in a circular orbit

B along a parabolic path

C along a straight line

D in an elliptical orbit.

Correct Answer

Option A

Solution

In magnetic field a charged particle moves in a circular orbit.

Q62

Monochromatic light of frequency 6.0

\times

10 14 Hz is produced by a laser. The power emitted is 2

\times

10

-

3 W. The number of photons emitted, on the average, by the source per second is

A 5

\times {10^{16}}

B 5

\times {10^{17}}

C 5

\times {10^{14}}

D 5

\times {10^{15}}

Correct Answer

Option D

Solution

Since p = nh $\nu$ $\Rightarrow$ n =

{p \over {h\nu }}

=

{{2 \times {{10}^{ - 3}}} \over {6.6 \times {{10}^{34}} \times 6 \times {{10}^{14}}}}

= 5

\times {10^{15}}

Q63

In a discharge tube ionization of enclosed gas is produced due to collisions between

A neutral gas atoms/molecules

B positive ions and neutral atoms/molecules

C negative electrons and neutral atoms/molecules

D photons and neutral atoms/molecules.

Correct Answer

Option C

Solution

In a discharge tube ionization of enclosed gas is produced due to collisions between negative electrons and neutral atoms/molecules.

Q64

When photons of energy h

\upsilon

fall on an aluminimum plate (of work function E 0 ), photoelectrons of maximum kinetic energy K are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

A K + h

\upsilon

B K + E 0

C 2K

D K.

Correct Answer

Option A

Solution

According to Einstein’s photoelectric equation, K = h $\nu$ – E 0 ... (i) and K' = h(2 $\nu$ ) – E 0 ... (ii) = 2h $\nu$ – E 0 = h $\nu$ + h $\nu$ – E 0 K'= h $\nu$ + K [using (i)]

Q65

A photocell employs photoelectric effect to convert

A change in the frequency of light into a change in the electric current

B change in the frequency of light into a change in electric voltage

C change in the intensity of illumination into a change in photoelectric current

D change in the intensity of illumination into a change in the work function of the photocathode.

Correct Answer

Option C

Solution

The photoelectric current is directly proportional to the intensity of illumination.

Therefore a change in the intensity of the incident radiation will change the photocurrent also.

Q66

The momentum of a photon of energy 1 MeV in kg m/s will be

A 5

\times

10

-

22

B 0.33

\times

10 6

C 7

\times

10

-

24

D 10

-

22

Correct Answer

Option A

Solution

Momentum of photon p = E/c $\therefore$ p =

{{1 \times {{10}^6} \times 1.6 \times {{10}^{19}}} \over {3 \times {{10}^8}}}

= 5 $\times$ 10 $-$ 22

Q67

A photosensitive metallic surface has work function, h

\upsilon

0 . If photons of energy

2h{\upsilon _0}

fall on this surface, the electrons come out with a maximum velocity of 4

\times

10 6 m/s. When the photon energy is increased to 5 h

\upsilon

0 , then maximum velocity of photoelectrons will be

A 2

\times

10 7 m/s

B 2

\times

10 6 m/s

C 8

\times

10 6 m/s

D 8

\times

10 5 m/s

Correct Answer

Option C

Solution

2h

{\nu _0}

= h

{\nu _0}

+

{1 \over 2}m{\left( {4 \times {{10}^6}} \right)^2}

5h

{\nu _0}

= h

{\nu _0}

+

{1 \over 2}m{v^2}

$\Rightarrow$

4 \times {1 \over 2}m{\left( {4 \times {{10}^6}} \right)^2}

=

{1 \over 2}m{v^2}

$\Rightarrow$ v = 8 $\times$ 10 6 m/s

Q68

The work functions for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein's equation the metals which will emit photoelectrons for a radiation of wavelength 4100

\mathop A\limits^ \circ

is/are

A A only

B A and B only

C all the three metals

D none

Correct Answer

Option B

Solution

Energy of photon =

{{12400} \over {4100}}

= 3 eV $\therefore$ A and B will emit photoelectrons.

Q69

J.J. Thomson's cathode-ray tube experiment demonstrated that

A cathode rays are streams of negatively charged ions

B all the maas of an atom is essentially in the nucleus

C the e/m of electrons is much greater than the e/m of protons

D the e/m ratio of the cathode-ray particles changes when a different gas is placed in the discharge tube

Correct Answer

Option C

Solution

{\left( {{e \over m}} \right)_{proton}} = {1 \over {1837}}{\left( {{e \over m}} \right)_{electron}}

$\therefore$

{\left( {{e \over m}} \right)_{proton}} \ll {\left( {{e \over m}} \right)_{electron}}

Q70

A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m then

A each emitted electron carries one quarter of the initial energy

B number of electrons emitted is half the initial number

C each emitted electron carries half the initial energy

D number of electrons emitted is a quarter of the initial number

Correct Answer

Option D

Solution

For a light source of power P watt, the intensity at a distance d is given by I =

{P \over {4\pi {d^2}}}

$\therefore$ Intensity $\propto$

{1 \over {{{\left( {dis\tan ce} \right)}^2}}}