JEE Advanced (Single Correct MCQs): Modern Physics - JEE MCQ

# JEE Advanced (Single Correct MCQs): Modern Physics - JEE MCQ

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## 30 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced (Single Correct MCQs): Modern Physics

JEE Advanced (Single Correct MCQs): Modern Physics for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The JEE Advanced (Single Correct MCQs): Modern Physics questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced (Single Correct MCQs): Modern Physics MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced (Single Correct MCQs): Modern Physics below.
JEE Advanced (Single Correct MCQs): Modern Physics - Question 1

### The plate resistance of a triode is 3 × 103 ohms and its mutual conductance is 1.5 × 10–3 amp/volt. The amplification factor of the triode is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 1

KEY CONCEPT : We know that µ = gm × r0
where µ = amplification factor,
gm = mutual conductance
r0 = plate resistance
∴ µ = 3 × 103 × 1.5 × 10–3 = 4.5

JEE Advanced (Single Correct MCQs): Modern Physics - Question 2

### The half life of radioactive Radon is 3.8 days. The time at the end of which 1/20 th of the radon sample will remainundecayed is (given log10 e = 0.4343)

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 2

t1/2 = 3.8 day

If the initial number of atom is a = A0 then after time t the number of atoms is a/20 = A. We have to find t.

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JEE Advanced (Single Correct MCQs): Modern Physics - Question 3

### An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 3

One point charge is  uranium nucleus

The other point charge is α particle  ∴ q= + 2e
Here the loss in K.E. = Gain in P.E. (till α-particle reaches the distance d)

= 529.92 × 10–16 m
= 529.92 × 10–14 cm
= 5.2992 × 10–12 cm

JEE Advanced (Single Correct MCQs): Modern Physics - Question 4

Beta rays emitted by a radioactive material are

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 4

β-particles are charged particles emitted by the nucleus.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 5

If elements with principal quantum number n > 4 were not allowed in nature, the  number of possible elements would be

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 5

KEY CONCEPT : The maximum number of electrons in an orbit is 2n2. n > 4 is not allowed.
Therefore the number of maximum electron that can be in first four orbits are 2 (1)2 + 2 (2)2 + 2 (3)2 + 2 (4)2
= 2 + 8 + 18 + 32 = 60
Therefore, possible element are 60.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 6

Consider the spectral line resulting from the  transition n = 2 → n = 1 in the atoms and ions given below. The shortest wavelength is produced by

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 6

We know that

λ is shortest when 1/λ is largest i.e., when Z has a highervalue. Z is highest for lithium.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 7

The equation  represents

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 7

JEE Advanced (Single Correct MCQs): Modern Physics - Question 8

Fast neutrons can easily be slowed down by

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 8

Fast neutrons can be easily slowed down by passing them through water.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 9

Consider α particles, β particles and γ - rays, each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are:

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 9

Note : The penetrating power is dependent on velocity.
For a given energy, the velocity of γ radiation is highest and α-particle is least.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 10

An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy in (eV) required to remove both the electrons from a neutral helium atom is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 10

When one e is removed from neutral helium atom, it becomes a one e species.
For one e species we know

For helium ion, Z = 2 and for first orbit n = 1.

∴ Energy required to remove this e  = + 54.4 eV

∴ Total energy required = 54.4 + 24.6 = 79 eV

JEE Advanced (Single Correct MCQs): Modern Physics - Question 11

A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years.
The time, in years, after which one-fourth of the material remains is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 11

when No is initial number of atoms

JEE Advanced (Single Correct MCQs): Modern Physics - Question 12

The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 12

KEY CONCEPT : For a semi conductor n = n0e–Eg/kT where n0 = no. of free electrons at absolute zero, n = no. of free electrons at T kelvin, Eg = Energy gap, k = Boltzmann constant.
As Eg increases, n decreases exponentially.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 13

A full-wave rectifier circuit along with the out-put is shown in Figure. The contribution (s) from the diode 1 is (are)

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 13

As shown in the fig. (i) during one half cycle the polarity of P and S are opposite such that diode (1) is reversed biased and hence non conducting.

During the other half cycle, diode (1) gets forward biased and is conducting. Thus diode (1) conducts in one half cycle and does not conduct in the other so the correct option is (b) (a and c.)

JEE Advanced (Single Correct MCQs): Modern Physics - Question 14

As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 14

KEY CONCEPT :

Therefore, ground state energy of doubly ionized lithium atom (Z = 3, n = 1) will be

∴ Ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4eV.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 15

The circuit shown in the figure contains two diodes each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6V, the current through the 100 ohm resistance (in  amperes) is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 15

In the circuit, diode D1 is forward biased, while D2 is reverse biased. Therefore, current i (through D1 and 100Ω resistance) will be

Here, 50Ω is the resistance of D1 in forward biasing.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 16

Which of the following statements is not true?

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 16

In n-type semiconductors, electrons are the majority charge carriers.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 17

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential, in volt, is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 17

Stopping potential is the negative potential applied to stop the electrons having maximum kinetic energy.
Therefore, stopping potential will be 4 volt.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 18

In hydrogen spectrum the wavelength of Hα line is  656 nm, whereas in the spectrum  of a distant galaxy, Hα line wavelength is 706 nm. Estimated speed of the galaxy with respect to earth is,

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 18

KEY CONCEPT : According to Doppler’s effect of light, the wavelength shift is given by

JEE Advanced (Single Correct MCQs): Modern Physics - Question 19

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, λ12, is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 19

Applying conservation of linear momentum, Initial momentum = Final momentum

0 = m1v1 – m2v2 ⇒ m1v1 = m2v2

JEE Advanced (Single Correct MCQs): Modern Physics - Question 20

Which of the following is a correct statement?

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 20

Beta rays are same as cathode rays as both are stream of electrons.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 21

Order of magnitude of density of uranium nucleus is, [mp = 1.67 × 10–27kg]

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 21

Nuclear density of an atom of mass number A,

JEE Advanced (Single Correct MCQs): Modern Physics - Question 22

22Ne nucleus, after absorbing energy, decays into two aparticles and an unknown nucleus. The unknown nucleus is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 22

The new element X has atomic number 6. Therefore, it is carbon atom.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 23

Binding energy per nucleon vs mass number curve for nuclei is shown in the Figure. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 23

KEY CONCEPT : Energy is released when stability increases. This will happen when binding energy per nucleon increases.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 24

Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength λ (given in terms of the Rydberg constant R for the hydrogen atom) equal to

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 24

KEY CONCEPT :

For ordinary hydrogen atom, longest wavelength

With hypothetical particle, required wavelength

JEE Advanced (Single Correct MCQs): Modern Physics - Question 25

The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true ?

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 25

NOTE : As the electron comes nearer to the nucleus the potential energy decreases

JEE Advanced (Single Correct MCQs): Modern Physics - Question 26

Two radioactive materials X1 and X2 have decay constants 10λ and λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 26

JEE Advanced (Single Correct MCQs): Modern Physics - Question 27

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

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 27

KEY CONCEPT :

Energy of incident electrons is greater than the ionization energy of electrons in K-shell, the K-shell electrons will be knocked off. Hence, characteristic X-ray spectrum will be obtained.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 28

The electron emitted in beta radiation originates from

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 28

Note : In a nucleus neutron converts into proton as follows n → p+ + e–1 Thus, decay of neutron is responsible for b-radiation origination

JEE Advanced (Single Correct MCQs): Modern Physics - Question 29

The transition from the state n = 4 to n = 3 in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 29

For 2 to 1, 3 to 2 and 4 to 2 we get energy that  n = 4 to n = 3,

I.R. radiation has less energy than U.V. radiation.

JEE Advanced (Single Correct MCQs): Modern Physics - Question 30

The intensity of X-rays from a Coolidge tube is plotted against wavelength λ as shown in the figure. The minimum wavelength found is λC and the wavelength of the Kα line is λK. As the accelerating voltage is increased

Detailed Solution for JEE Advanced (Single Correct MCQs): Modern Physics - Question 30

KEY CONCEPT :

In case of Coolidge tube

Thus the cut off wavelength is inversely proportional to accelerating voltage. As V increases, λc decreases. λk is the wavelength of K line which is a characteristic of an atom and does not depend on accelerating voltage of bombarding electron since λk always refers to a photon wavelength of transition of e from the target element from 2 → 1.
The above two facts lead to the conclusion that λk – λc increases as accelerating voltage is increased.

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