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12th Grade > Physics

X - RAYS MCQs

Total Questions : 12 | Page 1 of 2 pages
Question 1. The intensity of x-rays from Coolidge tube is plotted against wavelength λ  for a molybdenum target at an accelerating voltage of 35 kV. The minimum wavelength found is λ0 and the wavelength of the Kα line is λKα. As the  accelerating voltage is increased from 35 kV to 70 kV -
The Intensity Of X-rays From Coolidge Tube Is Plotted Agains...
  1.    (λKα−λ0) increases
  2.    (λKα−λ0) decreases
  3.    λKα increases.
  4.    λKα decreases.
 Discuss Question
Answer: Option A. -> (λKα−λ0) increases
:
A
The Kα peak corresponds to radiation when the atom transitions between two energy levels due to an internal jump of an electron from the Lshell to the Kshell. It is characteristic for a particular element and will not depend on any external parameters, like the accelerating potential.
But increasing the accelerating potential from 35kV to 70kV will now allow x-rays carrying energies in the range 35-70 eV, which was not allowed earlier. There will now be a higher cut-off on the energy, and hence, a lower cutoff on the wavelength (since E1λ). The new cut-off wavelength λ0 will be lower than λ0, and farther from λKα.
The Intensity Of X-rays From Coolidge Tube Is Plotted Agains...
Thus, among the given options, the only thing that will change when you increase the accelerationg potential is that (λKαλ0) will increase.
Question 2. The wavelength λ of Kβ X-Ray is given by:
  1.    hcEK−EM
  2.    hcEK−EL
  3.    hcEL−EM
  4.    hcEL−EK
 Discuss Question
Answer: Option A. -> hcEK−EM
:
A
Kβ characteristic X-Ray phton is emitted when an electron from the M shell jumps to fill the electronic vacancy created in K shell.
Now, if we refer to the Energy Level Diagram, we find that The atomic energy when there is a vacancy in K shell is EK.
Also the atomic energy when the vacancy gets created in M shell is EM.
We see that EM < EK.
So the atom has lost this energy. How much ?
EKEM
This energy appears as the energy of Kβ characteristic X-Ray phton.
Hence hcλKβ=EKEM.
Which gives,
1λKβ=EKEM.
Question 3. The wavelength of K-alpha characteristic x-rays only depends upon: 
  1.    The atomic energy levels
  2.    Current in the Coolidge tube setup
  3.    Accelerating potential in the Coolidge tube
  4.    Can't comment
 Discuss Question
Answer: Option A. -> The atomic energy levels
:
A
Atomic energy levels decide the amount of energy released in an electronic transition, and the wavelength of a photon is dependent on their energy.
Question 4. The wavelength of Kα and Lα X-Rays of a material are 20 pm and 140 pm respectively. Find the wavelength of Kβ X-Ray of the material.
  1.    12.5 pm
  2.    15.5 pm
  3.    17.5 pm
  4.    19.5 pm
 Discuss Question
Answer: Option C. -> 17.5 pm
:
C
For the wavelength of Kα x-rays; hcλ1 = EKEL
for the wavelength of Lα x-rays; hcλ2 = ELEM
or the wavelength of Kβ x-rays; hcλ3 = EKEM
Now,
If we add the first two equation we get :
hc(1λ1+1λ2)=EKEM
Now from equations iii) and iv)
hcλ3=hc(1λ1+1λ2).
We get;
1λ3=(1λ1+1λ2).
Or,
1λ3=(120+1140)pm1
Or,
λ3=2800160pm.
Or,
λ3 = 17.5 pm
Question 5. Any energy level marked here in the diagram can best be:
Any Energy Level Marked Here In The Diagram Can Best Be: 
 
  1.    The energy of electron in that particular shell
  2.    The energy of atom when electron is removed from that particular shell
  3.    The energy of atom when electron is in that particular shell
  4.    It may be all of the above.
 Discuss Question
Answer: Option B. -> The energy of atom when electron is removed from that particular shell
:
B
Any Energy Level Marked Here In The Diagram Can Best Be: 
The energies marked in the diagram are the energies of the atom when a vacancy of electron is created in particular shells.
For example, the energy marked as EKis the energy of atom when one electron from the K shell is removed. And so on.
It is interesting to note that the energy of the atom with one vacancy in K shell is higher; and as we move towards the higher shells, the energies of the atom with vacancies in those shells decrease. Also, the energy is zero when the atom is in its ground state.
This can be understood simply, as the K shell electrons are bound to the nucleus most strongly; this force decreases as we move to higher shells. Hence, to remove an electron from the lowermost shell (K) maximum amount of energy is required to be supplied, this amount will decrease as we move towards higher shells; for the same reason.
In the ground state (no electrons removed) the atom’s energy will be the least, which is taken to be zero.
Question 6. The  Kα x-ray emission line of tungsten occurs at λ = 21 pm. What is the energy difference between EK and EL levels in this atom (MeV refers to mega electron volts;1MeV = 106 eV and hc = 1242 eVnm)?
  1.    0.59 MeV
  2.    1.2 MeV
  3.    59 KeV
  4.    13.6 eV
 Discuss Question
Answer: Option A. -> 0.59 MeV
:
A
Kαx-ray emission takes place when an electron jumps from L-shell to fill a vacancy created in K-shell.
Now EKis the energy of the atom when vacancy is in K-Shell, and ELIs the energy of the atom when vacancy is created in L-Shell. Also, EK >ELSo, when vacancy shifts from K to L, according to our agrument, the atomic energy decreases. How much ?
EKEL.
This difference results into X-ray photon.
Hence connecting the dots we can conclude that ; The energy difference between K and L levels is the energy of the photon.
Mathematically;
EKEL=Ephoton
Now,
EKEL=hcλ
Which gives after putting in the values;
EKEL=1242eVnm0.0021nm
Final calculations leads us to our answer;
EKEL=591428eV
Or,
EKEL=0.59MeV
Question 7. The Kα x-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K electron knocked out is 23.32 keV, what will be the energy of this atom when an L electron is knocked out? If required, take hc = 1242 eV.nm.
  1.    16.37 keV
  2.    5.82 keV
  3.    62.02 keV
  4.    9.45 keV
 Discuss Question
Answer: Option B. -> 5.82 keV
:
B
In our notation of energy levels, an energy EX will mean the energy of the atom when there's a missing electron in the Xshell ( X = K, L, M, N,....). Writing quantitatively.
E(Kα)
= EKEL
EL=EKE(Kα)
EL=EK(hcλKα)
EL=23.32keV(1242eV.nm0.071nm)=23.32keV17.5eV
EL=5.82keV.
Question 8. While playing with the Crook’s tube, what was Mr. Roentgen looking for ?
  1.    Discovery of Benzene
  2.    Behaviour of radiations being generated inside Crook’s tube
  3.    Speed of light
  4.    Discovery of fundamental particles
 Discuss Question
Answer: Option B. -> Behaviour of radiations being generated inside Crook’s tube
:
B
Sneaking through history, we understand that Mr. Roentgen was looking for the behaviour of the radiation being generated inside the Crook’s tube.
Question 9. X−KαY−KβZ−Kγ
XKα
YKβ
ZKγ
  1.     1-x, 2-y, 3-z
  2.    2-x, 1-y, 3-z
  3.    3-x. 2-y, 1-z
  4.     1-x, 3-y, 2-z
 Discuss Question
Answer: Option A. ->  1-x, 2-y, 3-z
:
A
This is a definition based question.
We defined as following :
Kα : When electronis transition takes place from L-shell to K-shell or vacancy shifts from K-shell to L-shell (this is what the energy level diagram indicates)
Kβ :When electronis transition takes place from M-shell to K-shell or vacancy shifts from K-shell to M-shell (this is what the energy level diagram indicates)
Kγ :When electronis transition takes place from N-shell to K-shell or vacancy shifts from K-shell to N-shell (this is what the energy level diagram indicates)
Got the answer!!
Question 10. Arrange the given atoms in ascending order of their energy.
1)
Arrange The Given Atoms In Ascending Order Of Their Energy.1...
II) 
Arrange The Given Atoms In Ascending Order Of Their Energy.1...
III)
Arrange The Given Atoms In Ascending Order Of Their Energy.1...
IV.
Arrange The Given Atoms In Ascending Order Of Their Energy.1...
 
  1.    I > II > III > IV
  2.    I < II < III < IV
  3.     I = II = III= IV
  4.     II > IV > III > I
 Discuss Question
Answer: Option D. ->  II > IV > III > I
:
D
It’s a very simple question. We have to compare the energy in all the four diagrams. Basically these diagrams represent an atom with different electronic configuration with, infact, an electron missing from a specific shell in each case. Proceeding organically, first of all let's say the energy in the first case (with each electron at its place) is x.
Now, in the second diagram the electron is removed from the K-shell. We need to supply energy to the atom to do this right? So let's say we supplied an energy ΔK.
So, the atom's energy now becomes x+ΔK.
Now in the third diagram the electron is removed from M shell. We need to supply energy even for that. Let's say that energy is ΔM.
So, atom’s energy becomes x+ΔM.
In the fourth diagram, similarly after removing electron from L-shell the atomic energy becomes x+ΔL.
Now, we understand that it is tougher to remove the electron which is closer to the nucleus as it is more strongly bounded with the nucleus. To remove the electron from K-shell we need to supply the highest amount of energy, and to remove an electron from the M-shell we need to supply the least amount of energy. Mathematically,ΔK >ΔL >ΔM
Hence the energy of the atom will be in the order: II > IV > III > I.

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