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Test: Dual Nature of Matter - NEET MCQ


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25 Questions MCQ Test Physics Class 12 - Test: Dual Nature of Matter

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Test: Dual Nature of Matter - Question 1

What is the de-Broglie wavelength associated with an electron accelerated through a potential of 100 volts?

Detailed Solution for Test: Dual Nature of Matter - Question 1

and as we know 1Å = 0.1 nm, so 1.227Å = 0.122 nm

Test: Dual Nature of Matter - Question 2

An α – particle and a deutron are accelerated through the same potential difference. What will be the ratio of their de-Broglie wavelength?​

Detailed Solution for Test: Dual Nature of Matter - Question 2

Using conservation of energy, change in electrostatic energy equals the kinetic energy gained.

De-Broglie wavelength is given by:

⇒ This is the ratio of wavelength of deutron to alpha particle.

So,  Ratio of wavelength of alpha particle to deutron will be 1/2.

Test: Dual Nature of Matter - Question 3

The work function of metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be

Detailed Solution for Test: Dual Nature of Matter - Question 3

Step 1: Calculate the energy of the incident photon (h * nu)

The energy of a photon is given by:

where:

  • h = Planck's constant = 6.63 x 10⁻³⁴ Js

  • c = speed of light = 3 x 10⁸ m/s

Substitute the values:
E = (6.63 x 10⁻³⁴ * 3 x 10⁸) / (3 x 10⁻⁷) = 6.63 x 10⁻¹⁹ J

Convert this energy to electron volts (eV) using 1 eV = 1.6 x 10⁻¹⁹ J:
E = (6.63 x 10⁻¹⁹) / (1.6 x 10⁻¹⁹) = 4.14 eV

Step 2: Calculate the kinetic energy of the photoelectron

Using the photoelectric equation:
KE = h * nu - phi = 4.14 eV - 1 eV = 3.14 eV

Convert KE to joules:
KE = 3.14 x 1.6 x 10⁻¹⁹ = 5.024 x 10⁻¹⁹ J

Step 3: Determine the velocity of the photoelectron

The kinetic energy is related to velocity (v) by:
KE = (1/2) * m * v²

where:

  • m = mass of an electron = 9.11 x 10⁻³¹ kg

Rearrange to solve for v:
v = sqrt((2 * KE) / m)

Substitute the values:
v = sqrt((2 * 5.024 x 10⁻¹⁹) / (9.11 x 10⁻³¹))
v = sqrt(1.1 x 10¹²) ≈ 1.05 x 10⁶ m/s

Final Answer:

The velocity of the emitted photoelectrons is approximately 1 x 10⁶ m/s, which corresponds to option:

d) 1 x 10⁶ m/s

Test: Dual Nature of Matter - Question 4

Of the following moving with the same momentum, the one which has the largest wavelength is​

Detailed Solution for Test: Dual Nature of Matter - Question 4

De Broglie wavelength = h/mv,
where h = plancks constant, mv = momentum.
As they move with same momentum, de Broglie wavelength remains constant for all.

Test: Dual Nature of Matter - Question 5

The work function of a metallic surface is 5.01 eV. The photo-electrons are emitted when light of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photo-electrons is [ h = 4.14 x 10-15 eV sec]

Detailed Solution for Test: Dual Nature of Matter - Question 5

Energy of incident light E = 12375/2000 = 6.18 eV
According to relation E = W0 + eV

Test: Dual Nature of Matter - Question 6

A particle of mass M at rest decays into two particles of masses m1 and m2, having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, l1/l2 is

Detailed Solution for Test: Dual Nature of Matter - Question 6

By law of conservation of momentum 

–ve sign indicates that both the particles are moving in opposite direction. Now de-Broglie wavelengths 

*Multiple options can be correct
Test: Dual Nature of Matter - Question 7

When photon of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.70 eV is TB = (TA - 1.50) eV. If the de-Broglie wavelength of these photoelectrons is lB = 2lA, then

Detailed Solution for Test: Dual Nature of Matter - Question 7

Test: Dual Nature of Matter - Question 8

The de-Broglie wavelength of an electron is 1.0 nm. What is the retarding potential required to stop it?​

Detailed Solution for Test: Dual Nature of Matter - Question 8

λ=h/P
Here P= √2mk
K=kinetic energy
λ=h/√2mk
By energy conversion
K=eVs
λ=h/√2meVs
√Vs=h/ λ√2me
Vs=h2/ λ2me
Vs=(6.626x10-34)2/(1x10-9)2x2x9.11x10-31xe
Where e=1.602x10-19
Vs=43.9x10-68/29.1x10-68
   =1.508v
Vs=1.5v

Test: Dual Nature of Matter - Question 9

In a photoemissive cell with executing wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be

Detailed Solution for Test: Dual Nature of Matter - Question 9

Test: Dual Nature of Matter - Question 10

Photoelectric emission is observed from a metallic surface for frequencies v1 and v2 of the incident light rays (v1 > v2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is

Detailed Solution for Test: Dual Nature of Matter - Question 10

*Multiple options can be correct
Test: Dual Nature of Matter - Question 11

An X-ray tube is operating at 50 kV and 20 mA. The target material of the tube has a mass of 1.0 kg and specific heat 495 J kg–1°C–1. One percent of the supplied electric power is converted into X-rays and the entire remaining energy goes into heating the target. Then

Detailed Solution for Test: Dual Nature of Matter - Question 11

P = VI = 50 × 103 × 20 × 10-3 = 1000 W
Power converted into heat = 990 W 
MSΔT = 990 ⇒ ΔT = 2oc/sec
Now hc/1min = eV ⇒ hc/eV = 0.248 × 10-10m

Test: Dual Nature of Matter - Question 12

Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K shell electrons of tungsten have ionization energy 72.5 keV. X-rays emitted by the tube contain only

Detailed Solution for Test: Dual Nature of Matter - Question 12

Minimum wavelength of continuous X-ray spectrum is given by

Test: Dual Nature of Matter - Question 13

The X-ray wavelength of Lα line of platinum (Z = 78) is 1.30 Å. The X-ray wavelength of Lα line of Molybdenum (Z = 42) is

Detailed Solution for Test: Dual Nature of Matter - Question 13

The wavelength of Lα line is given by

Test: Dual Nature of Matter - Question 14

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature 27°C and 127°C respectively is

Detailed Solution for Test: Dual Nature of Matter - Question 14

de-Broglie wavelength rms velocity of a gas particle at the given temperature (T) is given as

Test: Dual Nature of Matter - Question 15

A photon of wavelength 6630 Å is incident on a totally reflecting surface. The momentum delivered by the photon is equal to

Detailed Solution for Test: Dual Nature of Matter - Question 15

The momentum of the incident radiation is given as p = h/λ
When the light is totally reflected normal to the surface the direction of the ray is reversed.
That means it reverses the direction of it's momentum without changing it's magnitude

Test: Dual Nature of Matter - Question 16

The ratio of de-Broglie wavelength of α-particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal to each other assuming the field induction vector  is perpendicular to the velocity vectors of the α-particle and the proton is

Detailed Solution for Test: Dual Nature of Matter - Question 16

When a charged particle (charge q, mass m) enters perpendicularly in a magnetic field (B) than, radius of the path described by it r = mv/qB ⇒ qBr.
Also de-Broglie wavelength λ = h/mv

Test: Dual Nature of Matter - Question 17

The potential energy of a particle of mass m is given by λ1 and λ2 are the de-Broglie wavelength of the particle, when 0 ≤ x ≤ 1 and x > 1 respectively. If the total energy of particle is 2E0, then the ratio λ1will be

Detailed Solution for Test: Dual Nature of Matter - Question 17

Test: Dual Nature of Matter - Question 18

Rest mass energy of an electron is 0.51 MeV. If this electron is moving with a velocity 0.8 c (where c is velocity of light in vacuum), then kinetic energy of the electrostatic should be 

Detailed Solution for Test: Dual Nature of Matter - Question 18

Given m0c2 = 0.51 MeV and v = 0.8 c
K.E. of the electron = mc2 - m0c
but
Now
mc2 = 0.51/0.6meV = 0.85 MeV
∴ K.E. = (0.85 - 0.51) MeV = 0.34 MeV

Test: Dual Nature of Matter - Question 19

A proton, a deuteron and an a-particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is 

Detailed Solution for Test: Dual Nature of Matter - Question 19

The deflection suffered by charged particle in an electric field is

Test: Dual Nature of Matter - Question 20

In order to coincide the parabolas formed by singly ionized ions in one spectrograph and doubly ionized ions in the other Thomson's mass spectrograph, the electric fields and magnetic fields are kept in the ratios 1: 2 and 3: 2 respectively. Then the ratio of masses of the ions is

Detailed Solution for Test: Dual Nature of Matter - Question 20

Usingwhere For parabolas to coincide in the two photographs, the kq/m, should be same for the two cases. Thus,

Test: Dual Nature of Matter - Question 21

Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell the work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero the voltage of the anode relative to the cathode must be made

Detailed Solution for Test: Dual Nature of Matter - Question 21

E = W0 + eV
For hydrogen atom, E = +13.6 eV
∴ + 13.6 = 4.2 + eV

Potential at anode = –9.4 V 

Test: Dual Nature of Matter - Question 22

The largest distance between the interatomic planes of a crystal is 10–7 cm. The upper limit for the wavelength of Xrays which can be usefully studied with this crystal is

Detailed Solution for Test: Dual Nature of Matter - Question 22

Bragg's law, 2d sin θ = nλ or 
For maximum wavelength, nmin = 1, (sin θ)max = 1
∴ λ = 2d   or λmax = 2 × 10-7 cm = 20 Å

Test: Dual Nature of Matter - Question 23

The wavelength of Kα X-rays produced by an X-ray tube is 0.76 Å. The atomic number of the anode material of the tube is

Detailed Solution for Test: Dual Nature of Matter - Question 23



z ≈ 40

Test: Dual Nature of Matter - Question 24

The Kα X-ray emission line of tungsten occurs at λ = 0.021 nm. The energy difference between K and L levels in this atom is about

Detailed Solution for Test: Dual Nature of Matter - Question 24

Line Kα corresponds to transfer of electron from L-shell to K-shell
Here, λ = 0.021nm = 2.1 × 10−11m
By the relation, E = hc/λ or

= 5.9 × 104eV = 59 KeV

Test: Dual Nature of Matter - Question 25

A silver ball of radius 4.8 cm is suspended by a thread in the vacuum chamber. UV light of wavelength 200 nm is incident on the ball for some times during which a total energy of 1 × 10–7 J falls on the surface. Assuming on an average one out of 103 photons incident is able to eject electron. The potential on sphere will be

Detailed Solution for Test: Dual Nature of Matter - Question 25

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