Modern Physics Test

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Modern Physics Test - 1

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Question 1
4 / -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

A
5 × 10^–5

B
4.5

C
45

D
2 × 10⁵

Solution

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
Question 2
4 / -1

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 log₁₀e = 0.4343)

A
3.8 days

B
16.5 days

C
33 days

D
76 days.

Solution

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.
Question 3
4 / -1

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

A
1 Å

B
10^–10 cm

C
10^–12cm

D
10^–15 cm

Solution

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
Question 4
4 / -1

Beta rays emitted by a radioactive material are

A
electromagnetic radiations

B
the electrons orbiting around the nucleus

C
charged particles emitted by the nucleus

D
neutral particles

Solution

β-particles are charged particles emitted by the nucleus.
Question 5
4 / -1

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

A
60

B
32

C
4

D
64

Solution

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.
Question 6
4 / -1

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

A
Hydrogen atom

B
Deuterium atom

C
Singly ionized Helium

D
Doubly ionised Lithium

Solution

We know that
λ is shortest when 1/λ is largest i.e., when Z has a highervalue. Z is highest for lithium.
Question 7
4 / -1

The equation represents

A
β -decay

B
γ-decay

C
fusion

D
fission

Solution
Question 8
4 / -1

Fast neutrons can easily be slowed down by

A
the use of lead shielding

B
passing them through water

C
elastic collisions with heavy nuclei

D
applying a strong electric field.

Solution

Fast neutrons can be easily slowed down by passing them through water.
Question 9
4 / -1

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:

A
α, β, γ

B
α, γ, β

C
β, γ,α

D
γ, β, α

Solution

Note : The penetrating power is dependent on velocity.
For a given energy, the velocity of γ radiation is highest and α-particle is least.
Question 10
4 / -1

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

A
38.2

B
49.2

C
51.8

D
79.0

Solution

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