JEE Advanced Previous Year Questions (2018 - 2024): Dual Nature of Radiation | Physics for JEE Main & Advanced PDF Download

 Page 1


 
Q1: A metal target with atomic number Z = 46 is bombarded with a high-energy electron beam.
The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K a-line
and the cut-off is found to be r = 2. If the same electron beam bombards another metal target
with Z = 41, the value of r will be:
(a) 2.53
(b) 1.27
(c) 2.24
(d) 1.58    [JEE Advanced 2024 Paper 2]
Ans: (a)
T o solve this problem, we need to understand the relationship between the atomic number Z, the
characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.
The cut-off wavelength, also known as the minimum wavelength ( ? ), corresponds to the
maximum energy of the X-rays produced and can be related to the accelerating voltage used in the
X-ray tube. The characteristic X-rays, such as the K -line, correspond to specic electron transitions
in the atom and are dependent on the atomic number Z of the target material.
The wavelength of the K -line is given by Moseley’s Law, which can be approximated as:
? ? 
1
(Z - 1)
The cut-off wavelength is given by the relationship between the energy of the incident electrons and
the X-ray produced:
? = 
hc
eV
Now, given the ratio r for the rst metal target with Z = 46:
r = 
?
?
 = 2
min
a
a
K a
2
min
K a
min
Page 2


 
Q1: A metal target with atomic number Z = 46 is bombarded with a high-energy electron beam.
The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K a-line
and the cut-off is found to be r = 2. If the same electron beam bombards another metal target
with Z = 41, the value of r will be:
(a) 2.53
(b) 1.27
(c) 2.24
(d) 1.58    [JEE Advanced 2024 Paper 2]
Ans: (a)
T o solve this problem, we need to understand the relationship between the atomic number Z, the
characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.
The cut-off wavelength, also known as the minimum wavelength ( ? ), corresponds to the
maximum energy of the X-rays produced and can be related to the accelerating voltage used in the
X-ray tube. The characteristic X-rays, such as the K -line, correspond to specic electron transitions
in the atom and are dependent on the atomic number Z of the target material.
The wavelength of the K -line is given by Moseley’s Law, which can be approximated as:
? ? 
1
(Z - 1)
The cut-off wavelength is given by the relationship between the energy of the incident electrons and
the X-ray produced:
? = 
hc
eV
Now, given the ratio r for the rst metal target with Z = 46:
r = 
?
?
 = 2
min
a
a
K a
2
min
K a
min
For the second metal target with Z = 41, let's denote the new ratio as r . We assume the same cutoff
wavelength since the same electron beam is used.
The ratio for Z = 41 can be calculated using the relationship between the K -line wavelengths:
?
?
 = 
(Z - 1)
(Z - 1)
 
Substituting Z = 46 and Z = 41:
?
?
 = 
(41 - 1)
(46 - 1)
  = 
40
45
  = 
8
9
  = 
64
81
Let ? = 2 ? because r = 2. Therefore:
? = ? · 
81
64
Thus, the new ratio r is:
r = 
?
?
 = 2 · 
81
64
 = 
162
64
 = 2.53
So, the value of r for the second metal target with Z = 41 is:
Option A: 2.53
2
a
K a(1)
K a(2)
2
1
2
1 2
K a(1)
K a(2)
2 2 2
K a(1) min
K a(2) K a(1)
2
2
K a(2)
min
Page 3


 
Q1: A metal target with atomic number Z = 46 is bombarded with a high-energy electron beam.
The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K a-line
and the cut-off is found to be r = 2. If the same electron beam bombards another metal target
with Z = 41, the value of r will be:
(a) 2.53
(b) 1.27
(c) 2.24
(d) 1.58    [JEE Advanced 2024 Paper 2]
Ans: (a)
T o solve this problem, we need to understand the relationship between the atomic number Z, the
characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.
The cut-off wavelength, also known as the minimum wavelength ( ? ), corresponds to the
maximum energy of the X-rays produced and can be related to the accelerating voltage used in the
X-ray tube. The characteristic X-rays, such as the K -line, correspond to specic electron transitions
in the atom and are dependent on the atomic number Z of the target material.
The wavelength of the K -line is given by Moseley’s Law, which can be approximated as:
? ? 
1
(Z - 1)
The cut-off wavelength is given by the relationship between the energy of the incident electrons and
the X-ray produced:
? = 
hc
eV
Now, given the ratio r for the rst metal target with Z = 46:
r = 
?
?
 = 2
min
a
a
K a
2
min
K a
min
For the second metal target with Z = 41, let's denote the new ratio as r . We assume the same cutoff
wavelength since the same electron beam is used.
The ratio for Z = 41 can be calculated using the relationship between the K -line wavelengths:
?
?
 = 
(Z - 1)
(Z - 1)
 
Substituting Z = 46 and Z = 41:
?
?
 = 
(41 - 1)
(46 - 1)
  = 
40
45
  = 
8
9
  = 
64
81
Let ? = 2 ? because r = 2. Therefore:
? = ? · 
81
64
Thus, the new ratio r is:
r = 
?
?
 = 2 · 
81
64
 = 
162
64
 = 2.53
So, the value of r for the second metal target with Z = 41 is:
Option A: 2.53
2
a
K a(1)
K a(2)
2
1
2
1 2
K a(1)
K a(2)
2 2 2
K a(1) min
K a(2) K a(1)
2
2
K a(2)
min
  
 
 
Q1: In a radioactive decay process, the activity is defined as ?? = -
????
????
, where ?? ( ?? ) is the number of 
radioactive nuclei at time ?? . Two radioactive sources, ?? ?? and ?? ?? have same activity at time ?? = ?? . At a 
later time, the activities of ?? ?? and ?? ?? are ?? ?? and ?? ?? , respectively. When ?? ?? and ?? ?? have just 
completed their ?? rd 
 and ?? th 
 half-lives, respectively, the ratio ?? ?? / ?? ?? is             [JEE Advanced 2023 
Paper 2] 
Ans: 16 
?? 1
= ?? 0
?? - ?? 1
?? 1
 also ?? 2
= ?? 0
?? - ?? 2
?? 2
 
At ?? 1
=
3 l n ? 2
?? 1
, 
?? 1
= ?? 0
?? - ?? 1
3 l n ? 2
?? 1
? = ?? 0
?? - 3 l n ? 2
… … …
 
Similarly, at 
?? 2
=
7ln ? 2
?? 2
,
?? 2
= ?? 0
?? - ?? 2
7 l n ? 2
?? 2
? = ?? 0
?? - 7 l n ? 2
… ( i )
 
Similarly, at 
?? 2
=
7ln ? 2
?? 2
,
?? 2
= ?? 0
?? - ?? 2
7 l n ? 2
?? 2
? = ?? 0
?? - 7 l n ? 2
…
 
From (i) and (ii) 
Page 4


 
Q1: A metal target with atomic number Z = 46 is bombarded with a high-energy electron beam.
The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K a-line
and the cut-off is found to be r = 2. If the same electron beam bombards another metal target
with Z = 41, the value of r will be:
(a) 2.53
(b) 1.27
(c) 2.24
(d) 1.58    [JEE Advanced 2024 Paper 2]
Ans: (a)
T o solve this problem, we need to understand the relationship between the atomic number Z, the
characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.
The cut-off wavelength, also known as the minimum wavelength ( ? ), corresponds to the
maximum energy of the X-rays produced and can be related to the accelerating voltage used in the
X-ray tube. The characteristic X-rays, such as the K -line, correspond to specic electron transitions
in the atom and are dependent on the atomic number Z of the target material.
The wavelength of the K -line is given by Moseley’s Law, which can be approximated as:
? ? 
1
(Z - 1)
The cut-off wavelength is given by the relationship between the energy of the incident electrons and
the X-ray produced:
? = 
hc
eV
Now, given the ratio r for the rst metal target with Z = 46:
r = 
?
?
 = 2
min
a
a
K a
2
min
K a
min
For the second metal target with Z = 41, let's denote the new ratio as r . We assume the same cutoff
wavelength since the same electron beam is used.
The ratio for Z = 41 can be calculated using the relationship between the K -line wavelengths:
?
?
 = 
(Z - 1)
(Z - 1)
 
Substituting Z = 46 and Z = 41:
?
?
 = 
(41 - 1)
(46 - 1)
  = 
40
45
  = 
8
9
  = 
64
81
Let ? = 2 ? because r = 2. Therefore:
? = ? · 
81
64
Thus, the new ratio r is:
r = 
?
?
 = 2 · 
81
64
 = 
162
64
 = 2.53
So, the value of r for the second metal target with Z = 41 is:
Option A: 2.53
2
a
K a(1)
K a(2)
2
1
2
1 2
K a(1)
K a(2)
2 2 2
K a(1) min
K a(2) K a(1)
2
2
K a(2)
min
  
 
 
Q1: In a radioactive decay process, the activity is defined as ?? = -
????
????
, where ?? ( ?? ) is the number of 
radioactive nuclei at time ?? . Two radioactive sources, ?? ?? and ?? ?? have same activity at time ?? = ?? . At a 
later time, the activities of ?? ?? and ?? ?? are ?? ?? and ?? ?? , respectively. When ?? ?? and ?? ?? have just 
completed their ?? rd 
 and ?? th 
 half-lives, respectively, the ratio ?? ?? / ?? ?? is             [JEE Advanced 2023 
Paper 2] 
Ans: 16 
?? 1
= ?? 0
?? - ?? 1
?? 1
 also ?? 2
= ?? 0
?? - ?? 2
?? 2
 
At ?? 1
=
3 l n ? 2
?? 1
, 
?? 1
= ?? 0
?? - ?? 1
3 l n ? 2
?? 1
? = ?? 0
?? - 3 l n ? 2
… … …
 
Similarly, at 
?? 2
=
7ln ? 2
?? 2
,
?? 2
= ?? 0
?? - ?? 2
7 l n ? 2
?? 2
? = ?? 0
?? - 7 l n ? 2
… ( i )
 
Similarly, at 
?? 2
=
7ln ? 2
?? 2
,
?? 2
= ?? 0
?? - ?? 2
7 l n ? 2
?? 2
? = ?? 0
?? - 7 l n ? 2
…
 
From (i) and (ii) 
?? 1
?? 2
=
?? 0
?? - 3 ln 2
?? 0
?? - 7 ln 2
=
2
- 3
2
- 7
=
1
2
- 4
= 2
4
= 16
? ?
?? 1
?? 2
= 16
 
Q2: A Hydrogen-like atom has atomic number ?? . Photons emitted in the electronic transitions from 
level ?? = ?? to level ?? = ?? in these atoms are used to perform photoelectric effect experiment on a 
target metal. The maximum kinetic energy of the photoelectrons generated is ?? . ?? ?? ???? . If the 
photoelectric threshold wavelength for the target metal is ?? ?? ?? ???? , the value of ?? is 
[Given: ???? = ?? ?? ?? ?? ???? - ???? and ?????? = ???? . ?? ???? , where ?? is the Rydberg constant, ?? is the Planck's 
constant and ?? is the speed of light in vacuum]         [JEE Advanced 2023 Paper 1] 
Ans: 3 
To find the atomic number ?? for the hydrogen-like atom emitting photons that cause photoelectrons to 
eject from the metal surface with a maximum kinetic energy of 1 . 95eV, we need to make use of the 
concept of energy transitions in atoms, as well as the photoelectric effect equation. We are given the 
photoelectric threshold wavelength for the metal and the constants h ?? and ?? h ?? . 
First, let's calculate the energy of the photon emitted during the transition from level ?? = 4 to level 
?? = 3 in the hydrogen-like atom. The energy of a photon emitted when an electron transitions between 
two levels in a hydrogen-like atom is given by : 
? ?? = ?? 2
?? h ?? (
1
?? 1
2
-
1
?? 2
2
) 
where: 
?? is the atomic number. 
?? h ?? is the ionization energy of hydrogen ( 13 . 6eV ) . 
?? 1
 and ?? 2
 are the principal quantum numbers of the initial and final energy levels, respectively. 
For the transition from ?? 1
= 4 to ?? 2
= 3, the energy of the photon is: 
For the transition from ?? 1
= 4 to ?? 2
= 3, the energy of the photon is: 
? ?? = ?? 2
· 13 . 6 · (
1
3
2
-
1
4
2
) eV
? ?? = ?? 2
· 13 . 6 · (
1
9
-
1
16
) eV
? ?? = ?? 2
· 13 . 6 · (
16 - 9
144
) eV
? ?? = ?? 2
· 13 . 6 ·
7
144
eV
 
Page 5


 
Q1: A metal target with atomic number Z = 46 is bombarded with a high-energy electron beam.
The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K a-line
and the cut-off is found to be r = 2. If the same electron beam bombards another metal target
with Z = 41, the value of r will be:
(a) 2.53
(b) 1.27
(c) 2.24
(d) 1.58    [JEE Advanced 2024 Paper 2]
Ans: (a)
T o solve this problem, we need to understand the relationship between the atomic number Z, the
characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.
The cut-off wavelength, also known as the minimum wavelength ( ? ), corresponds to the
maximum energy of the X-rays produced and can be related to the accelerating voltage used in the
X-ray tube. The characteristic X-rays, such as the K -line, correspond to specic electron transitions
in the atom and are dependent on the atomic number Z of the target material.
The wavelength of the K -line is given by Moseley’s Law, which can be approximated as:
? ? 
1
(Z - 1)
The cut-off wavelength is given by the relationship between the energy of the incident electrons and
the X-ray produced:
? = 
hc
eV
Now, given the ratio r for the rst metal target with Z = 46:
r = 
?
?
 = 2
min
a
a
K a
2
min
K a
min
For the second metal target with Z = 41, let's denote the new ratio as r . We assume the same cutoff
wavelength since the same electron beam is used.
The ratio for Z = 41 can be calculated using the relationship between the K -line wavelengths:
?
?
 = 
(Z - 1)
(Z - 1)
 
Substituting Z = 46 and Z = 41:
?
?
 = 
(41 - 1)
(46 - 1)
  = 
40
45
  = 
8
9
  = 
64
81
Let ? = 2 ? because r = 2. Therefore:
? = ? · 
81
64
Thus, the new ratio r is:
r = 
?
?
 = 2 · 
81
64
 = 
162
64
 = 2.53
So, the value of r for the second metal target with Z = 41 is:
Option A: 2.53
2
a
K a(1)
K a(2)
2
1
2
1 2
K a(1)
K a(2)
2 2 2
K a(1) min
K a(2) K a(1)
2
2
K a(2)
min
  
 
 
Q1: In a radioactive decay process, the activity is defined as ?? = -
????
????
, where ?? ( ?? ) is the number of 
radioactive nuclei at time ?? . Two radioactive sources, ?? ?? and ?? ?? have same activity at time ?? = ?? . At a 
later time, the activities of ?? ?? and ?? ?? are ?? ?? and ?? ?? , respectively. When ?? ?? and ?? ?? have just 
completed their ?? rd 
 and ?? th 
 half-lives, respectively, the ratio ?? ?? / ?? ?? is             [JEE Advanced 2023 
Paper 2] 
Ans: 16 
?? 1
= ?? 0
?? - ?? 1
?? 1
 also ?? 2
= ?? 0
?? - ?? 2
?? 2
 
At ?? 1
=
3 l n ? 2
?? 1
, 
?? 1
= ?? 0
?? - ?? 1
3 l n ? 2
?? 1
? = ?? 0
?? - 3 l n ? 2
… … …
 
Similarly, at 
?? 2
=
7ln ? 2
?? 2
,
?? 2
= ?? 0
?? - ?? 2
7 l n ? 2
?? 2
? = ?? 0
?? - 7 l n ? 2
… ( i )
 
Similarly, at 
?? 2
=
7ln ? 2
?? 2
,
?? 2
= ?? 0
?? - ?? 2
7 l n ? 2
?? 2
? = ?? 0
?? - 7 l n ? 2
…
 
From (i) and (ii) 
?? 1
?? 2
=
?? 0
?? - 3 ln 2
?? 0
?? - 7 ln 2
=
2
- 3
2
- 7
=
1
2
- 4
= 2
4
= 16
? ?
?? 1
?? 2
= 16
 
Q2: A Hydrogen-like atom has atomic number ?? . Photons emitted in the electronic transitions from 
level ?? = ?? to level ?? = ?? in these atoms are used to perform photoelectric effect experiment on a 
target metal. The maximum kinetic energy of the photoelectrons generated is ?? . ?? ?? ???? . If the 
photoelectric threshold wavelength for the target metal is ?? ?? ?? ???? , the value of ?? is 
[Given: ???? = ?? ?? ?? ?? ???? - ???? and ?????? = ???? . ?? ???? , where ?? is the Rydberg constant, ?? is the Planck's 
constant and ?? is the speed of light in vacuum]         [JEE Advanced 2023 Paper 1] 
Ans: 3 
To find the atomic number ?? for the hydrogen-like atom emitting photons that cause photoelectrons to 
eject from the metal surface with a maximum kinetic energy of 1 . 95eV, we need to make use of the 
concept of energy transitions in atoms, as well as the photoelectric effect equation. We are given the 
photoelectric threshold wavelength for the metal and the constants h ?? and ?? h ?? . 
First, let's calculate the energy of the photon emitted during the transition from level ?? = 4 to level 
?? = 3 in the hydrogen-like atom. The energy of a photon emitted when an electron transitions between 
two levels in a hydrogen-like atom is given by : 
? ?? = ?? 2
?? h ?? (
1
?? 1
2
-
1
?? 2
2
) 
where: 
?? is the atomic number. 
?? h ?? is the ionization energy of hydrogen ( 13 . 6eV ) . 
?? 1
 and ?? 2
 are the principal quantum numbers of the initial and final energy levels, respectively. 
For the transition from ?? 1
= 4 to ?? 2
= 3, the energy of the photon is: 
For the transition from ?? 1
= 4 to ?? 2
= 3, the energy of the photon is: 
? ?? = ?? 2
· 13 . 6 · (
1
3
2
-
1
4
2
) eV
? ?? = ?? 2
· 13 . 6 · (
1
9
-
1
16
) eV
? ?? = ?? 2
· 13 . 6 · (
16 - 9
144
) eV
? ?? = ?? 2
· 13 . 6 ·
7
144
eV
 
Next, using the photoelectric effect equation, the maximum kinetic energy (K.E.) of the ejected 
photoelectrons is equal to the energy of the incident photon minus the work function ( ?? ) . The work 
function can be calculated using the given threshold wavelength : 
?? =
h ?? ?? 
Given: ?? = 310 nm h ?? = 1 2 4 0 e V · nm 
So the work function is: 
?? =
1240
310
e V 
?? = 4e V 
The maximum kinetic energy is given by : 
?? . ?? . = ? ?? - ?? 
We are given ?? . ?? . = 1 . 95eV, hence : 
1 . 95 = ?? 2
· 13 . 6 ·
7
144
- 4
? ? ?? 2
· 13 . 6 ·
7
144
= 1 . 95 + 4
? ? ?? 2
=
1 . 95 + 4
13 . 6 ·
7
144
? ? ?? 2
=
5 . 95 × 144
13 . 6 × 7
? ? ?? 2
=
856 . 8
95 . 2
? ? ?? 2
= 9
? ? ?? = v 9
? ? ?? = 3
 
2022 
Q1: The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction 
?
?? ????
 ?? + ?
?? ?? ???? ? ?
?? ?? ?? + ?
?? ????
?? in a laboratory frame is ?? (in ?? ???? . Assume that ?
?? ????
 ?? is at rest in the 
laboratory frame. The masses of ?
?? ????
 ?? , ?
?? ?? ???? , ?
?? ?? ?? and ?
?? ????
?? can be taken to be 
???? . ?????? ?? , ?? . ?????? ?? , ?? . ?? ?? ?? ?? and ???? . ?????? ?? , respectively, where ?? ?? = ?? ?? ?? ?? ???? ?? - ?? . The value of ?? is   
[JEE Advanced 2022 Paper 1] 
Ans: 2.30 to 2.35