JEE Advanced (Single Correct MCQs): Modern Physics - Airforce X Y / Indian Navy SSR MCQ

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

t_1/2 = 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.

One point charge is  uranium nucleus

The other point charge is α particle  ∴ q₂ = + 2_e
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

β-particles are charged particles emitted by the nucleus.

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.

We know that

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

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

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

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

when N_o is initial number of atoms

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, E_g = Energy gap, k = Boltzmann constant.
As E_g increases, n decreases exponentially.

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.)

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.

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.

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

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

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

Applying conservation of linear momentum, Initial momentum = Final momentum

0 = m₁v₁ – m₂v₂ ⇒ m₁v₁ = m₂v₂

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

Nuclear density of an atom of mass number A,

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

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

KEY CONCEPT :

For ordinary hydrogen atom, longest wavelength

With hypothetical particle, required wavelength

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

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.

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

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.

KEY CONCEPT :

In case of Coolidge tube

Thus the cut off wavelength is inversely proportional to accelerating voltage. As V increases, λ_c decreases. λ_k is the wavelength of K_∝ line which is a characteristic of an atom and does not depend on accelerating voltage of bombarding electron since λ_k 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 λ_k – λ_c increases as accelerating voltage is increased.