JEE Advanced Previous Year Questions (2018 - 2023): Dual Nature of Radiation | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Advanced Previous Year Questions 
(2018 - 2023): Dual Nature of Radiation 
 
2023 
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 2


JEE Advanced Previous Year Questions 
(2018 - 2023): Dual Nature of Radiation 
 
2023 
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 3


JEE Advanced Previous Year Questions 
(2018 - 2023): Dual Nature of Radiation 
 
2023 
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 
Page 4


JEE Advanced Previous Year Questions 
(2018 - 2023): Dual Nature of Radiation 
 
2023 
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 
Q ? = ( m
N
+ m
He
- m
H
- m
O
) × c
2
? = ( 16 . 006 + 4 . 003 - 1 . 008 - 19 . 003 ) × 9 3 0 M e V
? = - 1 . 8 6 M e V
? = 1 . 8 6 M e V energy absorbed. 
 
And, 
1
2
×
?? × 4 ?? 5 ?? × ?? 2
= max loss in kinetic energy 
?
1
2
?? ?? 2
? =
5
4
× ?? ? =
5
4
× ( 1 . 86 ) M e V
? = 2 . 3 2 5 M e V
 
Q2: When light of a given wavelength is incident on a metallic surface, the minimum potential needed 
to stop the emitted photoelectrons is ?? . ?? ?? . This potential drops to ?? . ?? ?? if another source with 
wavelength four times that of the first one and intensity half of the first one is used. What are the 
wavelength of the first source and the work function of the metal, respectively? [Take 
????
?? = ?? . ???? ×
????
- ?? ?? ?? ?? - ?? .]    [JEE Advanced 2022 Paper 2] 
(a) ?? . ???? × ????
- ?? ?? , ?? . ?? ?? ???? 
(b) ?? . ???? × ????
- ?? ?? , ?? . ?? ?? ???? 
(c) ?? . ???? × ????
- ?? ?? , ?? . ?? ?? ???? 
(d) ?? . ???? × ????
- ?? ?? , ?? . ?? ?? ???? 
Ans: (a) 
By Einstein's photoelectric equation, we have: 
KE = h ?? - ?? h ?? - ?? = 6eV
h ?? ?? - ?? = 6eV … … . . . (i) 
h ?? 4 ?? - ?? = 0 . 6eV
3 h ?? 4 ?? = 5 . 4eV
? ? ?? =
3 h ?? 5 . 4eV × 4
? ? ?? =
3 × 1 . 24 × 10
- 6
4 × 5 . 4
= 1 . 72 × ×
- 7
 m . .
 
 
 
Page 5


JEE Advanced Previous Year Questions 
(2018 - 2023): Dual Nature of Radiation 
 
2023 
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 
Q ? = ( m
N
+ m
He
- m
H
- m
O
) × c
2
? = ( 16 . 006 + 4 . 003 - 1 . 008 - 19 . 003 ) × 9 3 0 M e V
? = - 1 . 8 6 M e V
? = 1 . 8 6 M e V energy absorbed. 
 
And, 
1
2
×
?? × 4 ?? 5 ?? × ?? 2
= max loss in kinetic energy 
?
1
2
?? ?? 2
? =
5
4
× ?? ? =
5
4
× ( 1 . 86 ) M e V
? = 2 . 3 2 5 M e V
 
Q2: When light of a given wavelength is incident on a metallic surface, the minimum potential needed 
to stop the emitted photoelectrons is ?? . ?? ?? . This potential drops to ?? . ?? ?? if another source with 
wavelength four times that of the first one and intensity half of the first one is used. What are the 
wavelength of the first source and the work function of the metal, respectively? [Take 
????
?? = ?? . ???? ×
????
- ?? ?? ?? ?? - ?? .]    [JEE Advanced 2022 Paper 2] 
(a) ?? . ???? × ????
- ?? ?? , ?? . ?? ?? ???? 
(b) ?? . ???? × ????
- ?? ?? , ?? . ?? ?? ???? 
(c) ?? . ???? × ????
- ?? ?? , ?? . ?? ?? ???? 
(d) ?? . ???? × ????
- ?? ?? , ?? . ?? ?? ???? 
Ans: (a) 
By Einstein's photoelectric equation, we have: 
KE = h ?? - ?? h ?? - ?? = 6eV
h ?? ?? - ?? = 6eV … … . . . (i) 
h ?? 4 ?? - ?? = 0 . 6eV
3 h ?? 4 ?? = 5 . 4eV
? ? ?? =
3 h ?? 5 . 4eV × 4
? ? ?? =
3 × 1 . 24 × 10
- 6
4 × 5 . 4
= 1 . 72 × ×
- 7
 m . .
 
 
 
2021 
Q1: In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals 
?? , ?? and ?? are ?? ?? , ?? ?? and ?? ?? , respectively, and they are related by ?? ?? = ?? ?? ?? = ?? ?? ?? . In this 
experiment, the same source of monochromatic light is used for metals ?? and ?? while a different 
source of monochromatic light is used for the metal R. The work functions for metals ?? , ?? and ?? are 
?? . ?? ???? , ?? . ?? ???? and ?? . ?? ???? , respectively. The energy of the incident photon used for metal ?? , in eV, is   
[JEE Advanced 2021 Paper 2] 
Ans: 6 
EP
P
= 2 E
Q
= 2 E
R
= E (assume)  
For metal P and Q, 
?? 1
- 4 = ?? ?? = ?? … … .
?? 1
- 4 . 5 = ?? ?? = ?? / 2
 
Subtracting equation (2) from (1), we get 
?? / 2 = 0 . 5 
For metal R 
?? 2
- 5 . 5 = ?? ?? = ?? / 2 = 0 . 5 
? E
2
= 6 
Q2: A heavy nucleus ?? , at rest, undergoes fission ?? ? ?? + ?? , where ?? and ?? are two lighter nuclei. 
Let ?? = ?? ?? - ?? ?? - ?? ?? , where ?? ?? , ?? ?? and ?? ?? are the masses of ?? , ?? and ?? , respectively. ?? ?? and 
?? ?? are the kinetic energies of ?? and ?? , respectively. The speeds of ?? and ?? are ?? ?? and ?? ?? , 
respectively. If ?? is the speed of light, which of the following statement(s) is(are) correct? 
(a) ?? ?? + ?? ?? = ?? ?? ?? 
(b) ?? ?? = (
?? ?? ?? ?? + ?? ?? ) ?? ?? ?? 
(c)  
?? ?? ?? ?? =
?? ?? ?? ?? 
(d) The magnitude of momentum for ?? as well ?? is ?? v ?? ???? , where ?? =
?? ?? ?? ?? ( ?? ?? + ?? ?? )
   [JEE Advanced 2021 
Paper 2] 
Ans: (a), (c) & (d)