Practice Start Test

Dual Nature of Radiation and Matter

NEET Physics · 96 questions · Page 8 of 10 · Click an option or "Show Solution" to reveal answer

Q71
The value of Planck's constant is
A 6.63 × \times 10 −-34 J/sec.
B 6.63 × \times 10 −-34 kg-m 2 /sec
C 6.63 × \times 10 −-34 kg-m 2
D 6.63 × \times 10 −-34 J-sec.
Correct Answer
Option D
Solution

The value of Planck’s constant is 6.63 ×\times 10 −-34 J-sec.

Q72
When ultraviolet rays incident on metal plate then photoelectric effect does not occur, it occurs by incidence of
A infrared rays
B X-rays
C radio wave
D micro wave
Correct Answer
Option B
Solution

It occurs by incidence of X-rays.

Q73
Which of the following is not the property of cathode rays ?
A In produces heating effect
B It does not deflect in electric field
C It casts shadow
D It produces fluorscence.
Correct Answer
Option B
Solution

Cathode rays are basically negatively charged particles (electrons).

If the cathode rays are allowed to pass between two plates kept at a difference of potential, the rays are found to be deflected from the rectilinear path.

The direction of deflection shows that the rays carry negative charges.

Q74
If particles are moving with same velocity, then which has maximum de Broglie wavelength?
A proton
B α\alpha -particle
C neutron
D β\beta -particle
Correct Answer
Option D
Solution

de Broglie wavelength for a particle is given by λ\lambda =

hmv{h \over {mv}}

Since all the particles are moving with same velocity, the particle with least mass will have maximum de-Broglie wavelength.

Here β\beta-particles(electrons) has the lowest mass and therefore it has maximum wavelength.

Q75
Which one among the following shows particle nature of light?
A photo electric effect
B interference
C refraction
D polarization.
Correct Answer
Option A
Solution

photo electric effect shows particle nature of light.

Q76
A photo-cell is illuminated by a source of light, which is placed at a distance d from the cell. If the distance become d/2, then number of electrons emitted per second, will be
A remain same
B four times
C two times
D one-fourth
Correct Answer
Option B
Solution

As Intensity becomes 4 times. So number of electrons emitted per second also becomes 4 times.

Q77
In Thomson mass spectrograph E→⊥B→\overrightarrow E \bot \overrightarrow B then the velocity of electron beam will be
A ∣E→∣∣B→∣{{\left| {\overrightarrow E } \right|} \over {\left| {\overrightarrow B } \right|}}
B E→×B→\overrightarrow E \times \overrightarrow B
C ∣B→∣∣E→∣{{\left| {\overrightarrow B } \right|} \over {\left| {\overrightarrow E } \right|}}
D E→2B→2{{{{\overrightarrow E }^2}} \over {{{\overrightarrow B }^2}}}
Correct Answer
Option A
Solution

As eE = evB ∴\therefore v =

∣E→∣∣B→∣{{\left| {\overrightarrow E } \right|} \over {\left| {\overrightarrow B } \right|}}
Q78
Who evaluated the mass of electron infirectly with help of charge
A Thomson
B Millikan
C Rutherford
D Newton
Correct Answer
Option A
Solution

Thomson evaluated the mass of electron infirectly with help of charge.

Q79
By photoelectric effect, Einstein proved
A E = hυ\upsilon
B K.E. = 12{1 \over 2} mv 2
C E = mc 2
D E = −Rhc2n2{{ - Rh{c^2}} \over {{n^2}}}
Correct Answer
Option A
Solution

By photoelectric effect, Einstein proved E = h

υ\upsilon

.

Q80
The photoelectric threshold wavelength of silver is 3250 × \times 10 −-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × \times 10 −-10 m is {Given h = 4.14 × \times 10 −-15 eV s and c = 3 × \times 10 8 m s −-1 ]
A ≈ \approx 0.6 × \times 10 6 m s −-1
B ≈ \approx 61 × \times 10 3 m s −-1
C ≈ \approx 0.3 × \times 10 6 m s −-1
D ≈ \approx 6 × \times 10 5 m s −-1
Correct Answer
Option A
Solution

Minimum energy, E 0 =

123753250{{12375} \over {3250}}

eV ⇒\Rightarrow E 0 = 3.81 eV Ultraviolet wavelength, λ\lambda = 2536

Ao\mathop A\limits^o

∴\therefore Imparted energy, E =

123752536{{12375} \over {2536}}

= 4.88 eV Now, max kinetic energy,

12mv2{1 \over 2}m{v^2}

= E - E 0 ⇒\Rightarrow v =

2×1.07×1.6×10−199.1×10−31\sqrt {{{2 \times 1.07 \times 1.6 \times {{10}^{ - 19}}} \over {9.1 \times {{10}^{ - 31}}}}}

⇒\Rightarrow v = 0.61 × 10 6 m/s = 6.0 × 10 5 m/s

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