Practice Start Test

Dual Nature of Radiation and Matter

NEET Physics · 96 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
A source S 1 is producing, 10 15 photons per second of wavelength 5000 A∘\mathop A\limits^ \circ . Another source S 2 is producing 1.02 × \times 10 15 photons per second of wavelength 5100 A∘\mathop A\limits^ \circ . Then, (power of S 2 )/(power of S 1 ) is equal to
A 1.00
B 1.02
C 1.04
D 0.98
Correct Answer
Option A
Solution

Power of S 2 Power of S 1 =

n2[hcλ2]n1[hcλ1]{{{n_2}\left[ {{{hc} \over {{\lambda _2}}}} \right]} \over {{n_1}\left[ {{{hc} \over {{\lambda _1}}}} \right]}}

=

n2λ1n1λ2{{{n_2}{\lambda _1}} \over {{n_1}{\lambda _2}}}

=

1.02×1015×50001015×5100{{1.02 \times {{10}^{15}} \times 5000} \over {{{10}^{15}} \times 5100}}

= 1

Q52
The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work functions 5.01 eV, when ultraviolet light of 200 nm falls on it, must be
A 2.4 V
B −- 1.2 V
C −- 2.4 V
D 1.2 V
Correct Answer
Option B
Solution

According to Einstein’s photoelectric equation eV s = hν\nu - ϕ\phi ⇒\Rightarrow eV s =

hcλ{{hc} \over \lambda }

- ϕ\phi =

1240200{{1240} \over {200}}

- 5.01 = 6.2 eV – 5.01 eV = 1.2 eV Stopping potential, V s = 1.2 V The potential difference that must be applied to stop photoelectrons = – V s = – 1.2 V

Q53
A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by (Where V is the potential difference between cathode and anode)
A B22VE2{{{B^2}} \over {2V{E^2}}}
B 2VB2E2{{2V{B^2}} \over {{E^2}}}
C 2VE2B2{{2V{E^2}} \over {{B^2}}}
D E22VB2{{{E^2}} \over {2V{B^2}}}
Correct Answer
Option D
Solution

Force on electron due to = Force on electron due magnetic field to electric field Bev = eE ⇒\Rightarrow v =

EB{E \over B}

.......(1) If V is the potential difference between the anode and the cathode, then

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

= eV ⇒\Rightarrow

em=v22V{e \over m} = {{{v^2}} \over {2V}}

.....(2) Substituting the value of v from equation (1) in equation (2), we get

em=E22VB2{e \over m} = {{{E^2}} \over {2V{B^2}}}
Q54
Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per sec. on the average at a target irradiated by this beam is
A 3 × \times 10 16
B 9×10159 \times {10^{15}}
C 3×10193 \times {10^{19}}
D 9×10179 \times {10^{17}}
Correct Answer
Option A
Solution

Number of photons =

Ehcλ{E \over {{{hc} \over \lambda }}}

=

9×10−3×6.67×10−76.6×10−34×3×108{{9 \times {{10}^{ - 3}} \times 6.67 \times {{10}^{ - 7}}} \over {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}

= 3 ×\times 10 16

Q55
The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement?
A Curves (a) and (b) represent incident radiations of same frequency but of different intensities.
B Curves (b) and (c) represent incident radiations of different frquencies and different intensities.
C Curves (b) and (C) represent incident radiations of same frequency having same intensity.
D Curves (a) and (b) represent incident radiations of different frequencies and different intensities
Correct Answer
Option A
Solution

Retarding potential depends on the frequency of incident radiation but is independent of intensity.

Q56
The number of photo electrons emitted for light of a frequency υ\upsilon (higher than the threshold frequency υ0{\upsilon _0}) is proportional to
A threshold frequency (υ0)\left( {{\upsilon _0}} \right)
B intensity of light
C frequency of light (υ0)\left( {{\upsilon _0}} \right)
D υ−υ0\upsilon - {\upsilon _0}
Correct Answer
Option B
Solution

The number of photoelectrons emitted is proportional to the intensity of incident light.

Saturation current ∝\propto intensity.

Q57
A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 × \times 10 6 m s −-1 . The velocity of the particle is
A 3 × \times 10 −-31 ms −-1
B 2.7×10−21 ms−12.7 \times {10^{ - 21}}\,m{s^{ - 1}}
C 2.7 × \times 10 −-18 ms −-1
D 9 × \times 10 −-2 ms −-1
Correct Answer
Option C
Solution

de-Broglie wavelength associated with electron moving with velocity ν, λ\lambda =

hmv{h \over {mv}}

So, λ\lambda e =

h9.1×10−31×3×106{h \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^6}}}

Wavelength of particle of mass 1 mg moving with velocity ν. λ\lambda p =

h10−6×v{h \over {{{10}^{ - 6}} \times v}}

As given, λ\lambda e = λ\lambda p ⇒\Rightarrow

h9.1×10−31×3×106{h \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^6}}}

=

h10−6×v{h \over {{{10}^{ - 6}} \times v}}

⇒\Rightarrow v =

27.3×10−2510−6{{27.3 \times {{10}^{ - 25}}} \over {{{10}^{ - 6}}}}

=

2.7×10−18 ms−12.7 \times {10^{ - 18}}\,m{s^{ - 1}}
Q58
The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the
A infrared region
B X-ray region
C Ultraviolet region
D Visible region
Correct Answer
Option C
Solution

ϕ\phi = 6.2 eV. K max = 5 eV; ∴\therefore hν\nu = 11.2 eV = E ∴\therefore λ\lambda =

hcE{{hc} \over E}

=

1240011.2{{12400} \over {11.2}}

= 1107

Ao\mathop A\limits^o
Q59
In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of
A collisions between the charged particles emitted from the cathode and the atoms of the gas
B collision between different electrons of the atoms of the gas
C excitation of electrons in the atoms
D collision between the atoms of the gas
Correct Answer
Option A
Solution

In the phenomenon of electric discharge tube through gases at low pressure, the coloured glow in the tube appears as a result of collisions between the charged particles emitted from cathode and the atoms of the gas.

Q60
A 5 watt source emits monochromatic light of wavelength 5000 A∘\mathop A\limits^ \circ . When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of
A 8
B 16
C 2
D 4
Correct Answer
Option D
Solution

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

P4πd2{P \over {4\pi {d^2}}}

∴\therefore No. of photoelectrons or intensity ∝\propto

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