Practice Start Test

Dual Nature of Radiation and Matter

NEET Physics · 96 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is
A 4 × \times 10 15 Hz
B 5 × \times 10 15 Hz
C 1.6 × \times 10 15 Hz
D 2.5 × \times 10 15 Hz
Correct Answer
Option C
Solution

Energy released from first exited state to the ground state : E = (– 3.4) – (– 13.6) = 10.2 eV Work function ϕ\phi = E – 3.75 eV = (10.2 – 3.75) eV = 6.63 eV ⇒\Rightarrow h

ν0{\nu _0}

= 6.63 ⇒\Rightarrow

ν0{\nu _0}

=

6.63×1.6×10−196.63×10−34{{6.63 \times 1.6 \times {{10}^{ - 19}}} \over {6.63 \times {{10}^{ - 34}}}}

= 1.6 ×\times 10 15 Hz

Q42
A 200 W sodium street lamp emits yellow light of wavelength 0.6 μ\mu m. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
A 1.5 × \times 10 20
B 6 × \times 10 18
C 62 × \times 10 20
D 3 × \times 10 19
Correct Answer
Option A
Solution

Efficient power, P =

Nt×hcλ{N \over t} \times {{hc} \over \lambda }

⇒\Rightarrow

Nt{N \over t}

=

25100×200(6.6×10−34)(3×108)0.6×10−6{{{{25} \over {100}} \times 200} \over {{{\left( {6.6 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)} \over {0.6 \times {{10}^{ - 6}}}}}}

= 1.5 ×\times 10 20

Q43
The threshold frequency for a photosensitive metal is 3.3 × \times 10 14 Hz. If light of frequency 8.2 × \times 10 14 Hz is incident on this metal, the cut- off voltage for the photoelectron emission is nearly
A 1 V
B 2 V
C 3 V
D 5 V
Correct Answer
Option B
Solution

K.E. = hν\nu – hν\nu th = eV 0 (V 0 = cut off voltage) ⇒\Rightarrow V 0 =

he(8.2×1014−3.3×1014){h \over e}\left( {8.2 \times {{10}^{14}} - 3.3 \times {{10}^{14}}} \right)

=

6.6×10−341.6×10−19(8.2×1014−3.3×1014){{6.6 \times {{10}^{ - 34}}} \over {1.6 \times {{10}^{ - 19}}}}\left( {8.2 \times {{10}^{14}} - 3.3 \times {{10}^{14}}} \right)

≈\approx 2 V

Q44
Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would
A increase by 2 times
B decrease by 2 times
C decrease by 4 times
D increase by 4 times
Correct Answer
Option B
Solution

The de Broglie wavelength λ\lambda associated with the electrons is

λ=1.227V\lambda = {{1.227} \over {\sqrt V }}

where V is the accelerating potential in volts. ∴\therefore λ\lambda ∝\propto

1V{1 \over {\sqrt V }}

⇒\Rightarrow

λ1λ2=V2V1{{{\lambda _1}} \over {{\lambda _2}}} = \sqrt {{{{V_2}} \over {{V_1}}}}

=

100×10325×103\sqrt {{{100 \times {{10}^3}} \over {25 \times {{10}^3}}}}

= 2 ⇒\Rightarrow

λ2=λ12{\lambda _2} = {{{\lambda _1}} \over 2}
Q45
Photoelectric emission occurs only when the incident light has more than a certain minimum
A power
B wavelength
C intensity
D frequency
Correct Answer
Option D
Solution

According to Einstein’s photoelectric equation K max = hν\nu – hν\nu 0 Since K max is +ve, the photoelectric emission occurs only if hν\nu >> hν\nu 0 or ν\nu >> ν\nu 0 The photoelectic emission occurs only when the incident light has more than a certain minimum frequency.

This minimum frequency is called threshold frequency.

Q46
In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by
A increasing the potential difference between the anode and filament
B increasing the filament current
C decreasing the filament current
D decreasing the potential difference between the anode and flament
Correct Answer
Option A
Solution

Davisson and Germer experiment describes that velocity of electron emitted from electron gun increases using potential difference between anode and filament.

Q47
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be
A 1 : 4
B 1 : 2
C 1 : 1
D 1 : 5
Correct Answer
Option B
Solution

According to Einsten’s photoelectric effect, the K.E. of the radiated electrons K.E max = E – W

12mv12{1 \over 2}mv_1^2

= (1 - 0.5) eV = 0.5 eV

12mv22{1 \over 2}mv_2^2

= (2.5 - 0.5) eV = 2 eV

v1v2=0.52{{{v_1}} \over {{v_2}}} = \sqrt {{{0.5} \over 2}}

= 1 : 2

Q48
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
A 1.8 V
B 1.3 V
C 0.5 V
D 2.3 V
Correct Answer
Option C
Solution

Maximum K.E. = Stopping Potential So stopping potential will be 0.5 V

Q49
When monochromatic radiation of intensity II falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2II, the number of emitted electrons and their maximum kinetic energy are respectively
A N and 2T
B 2N and T
C 2N and 2T
D N and T
Correct Answer
Option B
Solution

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

Maximum kinetic energy is independent of intensity of incident light but depends upon the frequency of light.

Hence option (b) is correct

Q50
The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in n th state E n = −13.6n2eV{{ - 13.6} \over {{n^2}}}eV)
A 5.1 V
B 12.1 V
C 17.2 V
D 7 V
Correct Answer
Option D
Solution

E = hν\nu = E 3 - E 1 =

−13.632−(−13.612)- {{13.6} \over {{3^2}}} - \left( { - {{13.6} \over {{1^2}}}} \right)

= 12.1 eV Therefore, stopping potential eV 0 =

hνh\nu

- ϕ\phi = (12.1 – 5.1) eV V 0 = 7 V

Ready for a full NEET mock test? Timed · full syllabus · instant results
Take a Mock Test →