Modern Physics Test

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Modern Physics Test - 5

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Question 1
4 / -1

A 280 days old radioactive substance shows an activity of 6000 dps, 140 days later its activity becomes 3000 dps.
What was its initial activity?

A
20000 dps

B
24000 dps

C
12000 dps

D
6000 dps

Solution

In two half lives, the activity will remain 1/4 of its initial activity.
Question 2
4 / -1

A pr oton has kinetic ener gy E = 100 keV which is equal to that of a photon. The wavelength of photon is λ₂ and that of proton is λ₁. The ration of λ₂/λ₁ is proportional to

A
E²

B
E^1/2

C
E^–1

D
E^{–1 /2}

Solution

For photon,
For proton,
...(ii)
Question 3
4 / -1

K_α wavelength emitted by an atom of atomic number Z = 11 is λ. Find the atomic number for an atom that emits K_α radiation with wavelength 4λ.

A
Z = 6

B
Z = 4

C
Z = 11

D
Z = 44

Solution
Question 4
4 / -1

A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV.
After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector?

A
One photon of energy 10.2 eV and an electron of energy 1.4 eV

B
2 photon of energy of 1.4 eV

C
2 photon of energy 10.2 eV

D
One photon of energy 10.2 eV and another photon of 1.4 eV

Solution

Initially a photon of energy 10.2eV collides inelastically with a hydrogen atom in ground state. For hydrogen atom,
The electron of hydrogen atom will jump to second orbit after absorbing the photon of energy 10.2 eV. The electron jumps back to its original state in less than microsecond and release a photon of energy 10.2 eV.
Another photon of energy 15 eV strikes the hydrogen atom inelastically. This energy is sufficient to knock out the electron from the atom as ionisation energy is 13.6 eV. The remaining energy of 1.4 eV is left with electron as its kinetic energy.
Question 5
4 / -1

A beam of electron is used in an YDSE experiment. The slit width is d. When the velocity of electron is increased, then

A
no interference is observed

B
fringe width increases

C
fringe width decreases

D
fringe width remains same

Solution

Note : Since electron shows wave nature, it will show the phenomenon of interference.
When speed of electron increases, λ will decrease. The distance between two consecutive fringes
As λ decreases, β also decrease.
Question 6
4 / -1

If a star can convert all the He nuclei completely into oxygen nuclei, the energy released per oxygen nuclei is [Mass of He nucleus is 4.0026 amu and mass of Oxygen nucleus is 15.9994 amu]

A
7.6 MeV

B
56.12 MeV

C
10.24 MeV

D
23.9 MeV

Solution

B.E. = Δm × 931.5 MeV
= (4 × 4.0026 – 15.9994) × 931.5 = 10.24 MeV
Question 7
4 / -1

is a radioactive substance having half life of 4 days. Find the probability that a nucleus undergoes decay after two half lives

A
1

B
1/2

C
3/4

D
1/4

Solution

For a nucleus to disintegr ate in two half life, the probability is 3/4 as 75% of the nuclei will disintegrate in this time.
Question 8
4 / -1

In the options given below, let E denote the rest mass energy of a nucleus and n a neutron.The correct option is

A

B

C

D

Solution

Iodine and Yttrium are medium sized nuclei and therefore, have more binding energy per nucleon as compared to Uranium which has a big nuclei and less B.E./nucleon.
In other words, Iodine and Yttrium are more stable and therefore possess less energy and less rest mass. Also when Uranium nuclei explodes, it will convert into I and Y nuclei having kinetic energies.
Question 9
4 / -1

The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is

A
802 nm

B
823 nm

C
1882 nm

D
1648 nm

Solution

The smallest frequency and largest wavelength in ultraviolet region will be for transition of electron from orbit 2 to orbit 1.
The highest frequency and smallest wavelength for infrared region will be for transition of electron from ∞ to 3rd orbit.
Question 10
4 / -1

Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

A

B

C

D
λ₀ = l

Solution

The cut off wavelength is given by
...(i)
According to de Broglie equation
From (i) and (ii),