JEE Previous Year Questions (2021-22): Dual Nature of Radiation - Physics 35 Year Past year Papers JEE Main & Advanced

Q.1. Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a×10⁻² W/m². The value of 'a' will be _________. [JEE Main 2022]

Ans. 84
P′ = 10% of 110W
=10 / 100 × 110 W
=11 W

=
= 84 x 10^-2 W/ m²

Q.2. A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be _________ hours. [JEE Main 2022]

Ans. 15
A0 = 64x, where x is safe limit
x = 64x × 2 − ^n/T_1/2
⇒164 = 2^−n/T1/2
or n/T1/2 = 6
⇒ n = 6 × 150 minutes
= 15 hours

Q.3. Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below :
The binding energies per nucleon for  respectively. The energy released in this process is _______________ MeV. [JEE Main 2022]

Ans. 26
Energy released = Change in B.E.
(7.6 × 4) − [4 × 1.1] = 26 MeV

Q.4. When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v₂. If v₂ = x v₁, the value of x will be __________. [JEE Main 2022]

Ans. 2
Let us say that work function is ϕ 
⇒
and
From (1) and (2)

Q.5. The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 A^o is 0.42 V. If the threshold frequency is x × 10¹³ /s, where x is _________ (nearest integer). [JEE Main 2022]
(Given, speed light = 3 × 10⁸ m/s, Planck's constant = 6.63 × 10⁻³⁴ Js)

Ans. 35

Q.6. The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength:
[JEE Main 2022]
(a)
(b)
(c)
(d)

Ans. b

Q.7. Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted,  in the two cases respectively, will be
[JEE Main 2022]
(a) 1 : 2
(b) 1 : 3

(c) 2 : 3
(d) 3 : 2

Ans. c

Q.8. The half life period of a radioactive substance is 60 days. The time taken for 7/8^th of its original mass to disintegrate will be : [JEE Main 2022]
(a) 120 days
(b) 130 days
(c) 180 days
(d) 20 days

Ans. c

=180 days

Q.9. With reference to the observations in photo-electric effect, identify the correct statements from below : [JEE Main 2022]
(A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light.
(B) The value of saturation current increases on moving the source of light away from the metal surface.
(C) The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light.
(D) The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves.
(E) Existence of threshold wavelength can not be explained by wave nature of light/ electromagnetic waves.
Choose the correct answer from the options given below:
(a) (A) and (B) only
(b) (A) and (E) only
(c) (C) and (E) only
(d) (D) and (E) only

Ans. b
∵
⇒  varies linearly with frequency.
And, threshold wavelength can be explained by particle nature of light.

Q.10. A metal exposed to light of wavelength 800 nm and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The workfunction of the metal is : (Take hc =1230eV−nm ).
[JEE Main 2022]
(a) 1.537 eV
(b) 2.46 eV
(c) 0.615 eV
(d) 1.23 eV

Ans. c

Q.11. A source of monochromatic light liberates 9 × 10²⁰ photon per second with wavelength 600 nm when operated at 400 W. The number of photons emitted per second with wavelength of 800 nm by the source of monochromatic light operating at same power will be : [JEE Main 2022]
(a) 12 × 10²⁰
(b) 6 × 10²⁰
(c) 9 × 10²⁰
(d) 24 × 10²⁰

Ans. a
As we know

⇒

Q.12. Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments 'B' and 'C' of mass numbers 105 and 115. The binding energy of nucleons in 'B' and 'C' is 6.4 MeV per nucleon. The energy Q released per fission will be : 
[JEE Main 2022]
(a) 0.8 MeV
(b) 275 MeV
(c) 220 MeV
(d) 176 MeV

Ans. d
²²⁰A → ¹⁰⁵B + ¹¹⁵C
⇒ Q = [105 × 6.4 + 115 × 6.4] − [220 × 5.6] MeV
⇒ Q = 176 MeV

Q.13. The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :
[JEE Main 2022]
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 1 : 4

Ans. b

Q.14. Given below are two statements : [JEE Main 2022]
Statement I : Davisson-Germer experiment establishes the wave nature of electrons.
Statement II : If electrons have wave nature, they can interfere and show diffraction.
In the light of the above statements choose the correct answer from the option given below:
(a) Both Statement I and Statement II are true.
(b) Both Statement I and Statement II are false.
(c) Statement I is true but Statement II is false.
(d) Statement I is false but Statement II is true.

Ans. a
Davisson-Germer experiment is done and establishes the wave nature of electrons. Interference and diffraction establishes wave nature.

Q.15. How many alpha and beta particles are emitted when Uranium ₉₂U²³⁸ decays to lead 82Pb²⁰⁶ ? [JEE Main 2022]
(a) 3 alpha particles and 5 beta particles
(b) 6 alpha particles and 4 beta particles
(c) 4 alpha particles and 5 beta particles
(d) 8 alpha particles and 6 beta particles

Ans. d

238 = 206 + 4x + 0
⇒ 4x = 32 ⇒ x = 8
also, 92 = 82 + 2x − y
y = 82 + 16 − 92 = 6

Q.16. Given below are two statements : [JEE Main 2022]
Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E₁) to higher energy orbit (E₂), is given as hf = E₁ − E₂
Statement II : The jumping of electron from higher energy orbit (E2) to lower energy orbit (E₁) is associated with frequency of radiation given as f = (E₂ − E₁)/h
This condition is Bohr's frequency condition.
In the light of the above statements, choose the correct answer from the options given below:
(a) Both Statement I and Statement II are true.
(b) Both Statement I and Statement II are false.
(c) Statement I is correct but Statement II is false.
(d) Statement I is incorrect but Statement II is true.

Ans. d
Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.
Also, E₂ − E₁ is the energy of photon
⇒ E₂ − E₁ = hf
⇒

Q.17. The Q-value of a nuclear reaction and kinetic energy of the projectile particle, K_p are related as : [JEE Main 2022]
(a) Q = K_p
(b) (K_p + Q) < 0
(c) Q < K_p
(d) (K_p + Q) > 0

Ans. d
K_p > 0
If Q is released ⇒ Q > 0
⇒ K_p + Q > 0
Even the particle has to be given kinetic energy greater than magnitude of Q to maintain momentum conservation.
⇒ K + Q > 0

Q.18. The electric field at a point associated with a light wave is given by
E = 200 [sin (6 × 10¹⁵)t + sin (9 × 10¹⁵)t] Vm⁻¹
Given : h = 4.14 × 10⁻¹⁵ eVs
If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be [JEE Main 2022]
(a) 1.90 eV
(b) 3.27 eV
(c) 3.60 eV
(d) 3.42 eV

Ans. d
Frequencies of EM Waves =
Energy of one photon of these waves
=
and
= 3.95 eV and 5.93 eV
⇒ Energy of maximum energetic electrons
= 5.93 − 2.50 = 3.43 eV

Q.19. Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : [JEE Main 2022]
Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal.
Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal.
In the light of the above statements, choose the most appropriate answer from the options given below.

(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is not correct.
(d) A is not correct but R is correct.

Ans. b
When energy of incident radiation is equal to the work function of the metal, then the KE of photoelectrons would be zero. But this statement does not comment on the situation when energy is less than the work function.

Q.20. When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V. This potential drops to 0.6 V 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? [JEE Advance 2022]
[Take hc/e = 1.24× 10⁻⁶JmC⁻¹.]
(a) 1.72 × 10⁻⁷ m,1.20eV
(b) 1.72 × 10⁻⁷ m,5.60eV
(c) 3.78 × 10⁻⁷ m,5.60eV
(d) 3.78 × 10⁻⁷ m,1.20eV

Ans. a

Q.21. The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ₇¹⁶ N+ ₂⁴He→₁¹H+₈¹⁹O in a laboratory frame is n (in MeV. Assume that ₇¹⁶ N is at rest in the laboratory frame. The masses of ₇¹⁶ N,₂⁴He,₁¹H and ₈¹⁹O can be taken to be 16.006u, 4.003u, 1.008u and 19.003u, respectively, where 1u = 930MeVc−2. The value of n is ________ . 
[JEE Advance 2022]

Ans. 2.325

And,  = max loss in kinetic energy
⇒

Q.22. The half life period of radioactive element x is same as the mean life time of another radioactive element y. Initially they have the same number of atoms. Then : [JEE Main 2021]
(a) x-will decay faster than y.
(b) y-will decay faster than x.
(c) x and y have same decay rate initially and later on different decay rate.
(d) x and y decay at the same rate always.

Ans. b
Given, (τ_1/2)_x = (τ)y
Here, τ_1/2 = half-life period of radioactive element and τ = mean life period of radioactive element.
As we know the expression,
Half-life of the radioactive element x,

Mean life of the radioactive element y,

Substituting the values in Eq. (i), we get

Initially they have same number of atoms,
N_x = Ny = N₀
As we know that,
Activity, A = λN
As, λ_x < λ_y ⇒ A_x < A_y
Therefore, y will decay faster than x.

Q.23. The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is :
[me = mass of electron = 9 × 10⁻³¹ kg, h = Planck constant = 6.6 × 6.6 × 10⁻³⁴ Js, kB = Boltzmann constant = 1.38 × 10⁻²³ JK⁻¹] [JEE Main 2021]
(a) 6.26 nm
(b) 8.46 nm
(c) 2.26 nm
(d) 3.25 nm

Ans. a
Given, Planck's constant, h = 6.6 × 10−34 Js
Boltzmann constant, kB = 1.38 × 10−23 J/K
Mass of an electron, me = 9 × 10−31 kg
Temperature of an ideal gas, T = 300 K
As we know that, de-Broglie wavelength,

Here, E is the kinetic energy,

Substituting value of E in Eq. (i), we get

Substituting the given values in the above equation, we get

= 6.26 nm
∴ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.

Q.24. Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :- [JEE Main 2021]
(a)

(b)
(c)
(d)

Ans. c

Q.25. A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 × 10⁻³⁴ Js) [JEE Main 2021]
(a) 1.45 × 10¹⁶ MHz
(b) 0.19 × 10¹⁵ MHz
(c) 1.45 × 10⁹ MHz
(d) 9.0 × 10²⁷ MHz

Ans. c
For every large distance P.E. = 0
& total energy = 2.6 + 0 = 2.6 eV
Finally in first excited state of H atom total energy = −3.4 eV
Loss in total energy = 2.6 − (−3.4) = 6 eV
It is emitted as photon

= 1.45 × 10⁹ Hz

Q.26. A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm? [JEE Main 2021]
(a) 0.96 V
(b) 1.25 V
(c) 0.24 V
(d) 1.5 V

Ans. b

when λ_i = 670.5 nm ; V_o = 0.48
when λ_i = 474.6 nm ; V_o = ?
So,

(2) − (1)

V_o = 0.48 + 0.76
V_o = 1.24 V ≃ 1.25 V

Q.27. In a photoelectric experiment, increasing the intensity of incident light : [JEE Main 2021]
(a) increases the number of photons incident and also increases the K.E. of the ejected electrons
(b) increases the frequency of photons incident and increases the K.E. of the ejected electrons
(c) increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged
(d) increases the number of photons incident and the K.E. of the ejected electrons remains unchanged

Ans. d
→ Increasing intensity means number of incident photons are increased.
→ Kinetic energy of ejected electrons depend on the frequency of incident photons, not the intensity.

Q.28. The de-Broglie wavelength of a particle having kinetic energy E is λ. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value? [JEE Main 2021]
(a) 1/9E
(b) 7/9E
(c) E
(d) 16/9E

Ans. b

Extra energy given = 16/9E - E = 7/9E

Q.29. In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function ϕ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 × 10⁻³⁴ Js, c = 3 × 10⁸ ms⁻¹) [JEE Main 2021]
(a) 1.3 V
(b) 1.1 V
(c) 1.9 V
(d) 0.6 V

Ans. a

ΔV = V_s1−V_s2 = 1.93 − 0.6 = 1.33 V

Q.30. If 'f' denotes the ratio of the number of nuclei decayed (Nd) to the number of nuclei at t = 0 (N0) then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as : [JEE Main 2021]
[λ is the radioactive decay constant]
(a) − λ (1 − e−^λt)
(b) λ (1 − e^−λt)
(c) λe^−λt
(d) − λe^−λt

Ans. c
N = N₀e^−λt
N_d = N₀ − N
N_d = N₀ (1 − e^−λt)

Q.31. When radiation of wavelength λ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is : [JEE Main 2021]
(a) 2λ
(b) 4λ
(c) 8λ
(d) 6λ

Ans. b
V_s = hv − ϕ
4.8 = hc / λ − ϕ ..... (i)
1.6 = hc / 2λ − ϕ ..... (ii)
Using above equation (i) - (ii)

Put in equation (ii)
ϕ = 1.6
hc / λ_th = 1.6
λ_th = hc / 1.6
=

Q.32. The half-life of ¹⁹⁸Au is 3 days. If atomic weight of ¹⁹⁸Au is 198 g/mol then the activity of 2 mg of ¹⁹⁸Au is [in disintegration/second] : [JEE Main 2021]
(a) 2.67 × 10¹²
(b) 6.06 × 10¹⁸
(c) 32.36 × 10¹²
(d) 16.18 × 10¹²

Ans. d
A = λN

= 2.67 × 10⁻⁶ sec⁻¹ 
N = Number of atoms in 2 mg Au
=
= 6.06 × 10¹⁵ 
A = λN = 1.618 × 10¹³ = 16.18 × 10¹² dps

Q.33. Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as : [JEE Main 2021]
(where λ is the decay constant)
(a)
(b)
(c)
(d)

Ans. d

ln⁡(3/4) = −λt₁ ..... (i)
ln⁡(1/2) = −λt₂ ..... (i)
ln⁡(3/4) − ln⁡(1/2) = λ(t₂ − t₁) ....(i)

Q.34. An electron having de-Broglie wavelength λ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is : [JEE Main 2021]
(a) 0
(b)
(c)
(d)

Ans. c
λ = h / mv
kinetic energy,

Q.35. The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 × 10⁻⁴ T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron = 9.1 × 10⁻³¹ kg) [JEE Main 2021]
(a) 1.36 eV
(b) 1.88 eV
(c) 0.82 eV
(d) 0.16 eV

Ans. c
Energy of photon can be given as

where, n₁ = lower energy level and n₂ = higher energy level.
As per question, n₁ = 2, n₂ = 3
∴
=
We know that work-function is the minimum energy required to eject photoelectrons from metal surface.
For gold plate, it will be
ϕ = EP − KE_max .... (i)
[Given, B = 5 × 10⁻⁴ T, r = 7 mm = 7 × 10⁻³ m, q = 1.6 × 10⁻¹⁹ C and m = 9.1 × 10⁻³¹ kg] 
Therefore, velocity of photoelectrons will be

Kinetic energy will be
∴
Now, putting the values, in Eq. (i), we get
ϕ = (1.89 − 1.075) eV = 0.82 eV

Q.36. Two identical photocathodes receive the light of frequencies f1 and f2 respectively. If the velocities of the photo-electrons coming out are v1 and v2 respectively, then [JEE Main 2021]
(a)
(b)
(c)
(d)

Ans. d

Subtracting equation (1) by equation (2)

Q.37. The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation : [JEE Main 2021]
(a) Phase
(b) Frequency
(c) Amnplitude
(d) Intensity

Ans. b
Stopping potential depends on frequency, according to Einstein's photoelectric equation.
hv − hv₀ = eV
⇒

Q.38. The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be : [JEE Main 2021]
(a) 4.34 m/s
(b) 2.19 m/s
(c) 3.25 m/s
(d) 4.17 m/s

Ans. d
(ΔE) Releases when photon going from n = 5 to n = 1
ΔE = (13.6 − 0.54) eV = 13.06 eV.
P_i = P_f (By linear momentum conservation)


= 4.17 m/sec

Q.39. Given below are two statements : one is labeled as Assertion A and the other is labelled as Reason R. [JEE Main 2021]
Assertion A : An electron microscope can achieve better resolving power than an optical microscope.
Reason R : The de Broglie's wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light.
In the light of the above statements, choose the correct answer from the options given below:
(a) A is false but R is true.
(b) Both A and R are true but R is NOT the correct explanation of A.
(c) Both A and R are true and R is the correct explanation of A.
(d) A is true but R is false.

Ans. c
Resolution limit (Δθ) = 1.22λ / d
Resolution power = 1 / Resolution limit
∴ Resolution power ∝ 1 / λ
Since, wavelength of electron is much less than visible light, its resolving power will be much more.

Q.40. The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is : [JEE Main 2021]
(a) 400 nm
(b) 329 nm
(c) 309 nm
(d) 382 nm

Ans. d
From the photoelectric effect equation

so,

Subtract equation (i) from equation (ii)


λ₂ = 383.14
λ₂ ≃ 382 nm

Q.41. The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is : [JEE Main 2021]
(a) 194.8 nm
(b) 490.7 nm
(c) 913.3 nm
(d) 121.8 nm

Ans. d
ΔE = 10.2 eV
hc/λ = 10.2 eV

= 121.56 nm
≃ 121.8 nm

Q.42. An α particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are λ_α and _λp respectively. The ratio λ_p / λ_α is :
[JEE Main 2021]
(a) 8
(b) 2.8
(c) 7.8
(d) 3.8

Ans. b
We know,


= √8
= 2√2

Q.43. According to Bohr atom model, in which of the following transitions will the frequency be maximum? [JEE Main 2021]
(a) n = 2 to n = 1
(b) n = 3 to n = 2
(c) n = 4 to n = 3
(d) n = 5 to n = 4

Ans. a
Let, n_f, n_i be the final and initial orbit.
As we know that,

Now, checking for each option, we get

The option (b) has highest value.
Since, frequency, f = c/λ ⇒ f ∝ 1/λ
∴ Frequency will be maximum for transition n = 2 to n = 1.

Q.44. Match List - I with List - II. [JEE Main 2021]
Choose the correct answer from the options given below :
(a) (a)-(vi), (b)-(v), (c)-(i), (d)-(iv)
(b) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(c) (a)-(ii), (b)-(iv), (c)-(vi), (d)-(iii)
(d) (a)-(vi), (b)-(iv), (c)-(i), (d)-(v)

Ans. b
(a) Source of microwave frequency - (ii) Magnetron
(b) Source of infra red frequency - (iv) Vibration of atom and molecules
(c) Source of gamma ray - (i) Radio active decay of nucleus
(d) Source of X-ray - (iii) inner shell electron

Q.45. An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be : [JEE Main 2021]
(a) 10⁻² nm
(b) 10⁻¹ nm
(c) 10⁻³ nm
(d) 10⁻⁴ nm

Ans. c
Given, V = 1.24 million volt = 1.24 × 10⁶ volt
Since, energy (E) = eV
where, e is the charge of electron = 1.6 × 10⁻¹⁹ C
∴ E = 1.6 × 10⁻¹⁹ × 1.24 × 10⁶ ..... (i)
As we know that,
Energy of photon, E = hc / λ .... (ii)
Here, Planck's constant, h = 6.67 × 10⁻³⁴ J-s,
c = speed of light in free space, c = 3 × 10⁸ ms⁻¹
Equating Eqs. (i) and (ii), we get

=1.009 × 10⁻¹² ≃ 10−3 × 10⁻⁹
= 10⁻³ nm

Q.46. Given below are two statements : [JEE Main 2021]
Statement I : Two photons having equal linear momenta have equal wavelengths.
Statement II : If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.
(a) Statement I is false but Statement II is true
(b) Both Statement I and Statement II are false
(c) Both Statement I and Statement II are true
(d) Statement I is true but Statement II is false

Ans. d
As we know,
If linear momenta of two photons are equal, then their wavelengths is also equal.
Also, if the wavelength is decreased, then the momentum and energy of photon will increase.
Hence, option (d) is correct.

Q.47. The Kα X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be __________ keV. (Round off to the nearest integer)
[h = 4.14 × 10⁻¹⁵ eVs, c = 3 × 10⁸ ms⁻¹]  [JEE Main 2021]

Ans. 10

E_L = (27.5 − 17.5) keV
= 10 keV

Q.48. A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 Ω in the collector circuit is 0.6 V. If the current gain factor (β) is 24, then the base current is _____________ μA. (Round off to the Nearest Integer) [JEE Main 2021]

Ans. 25

R_C = 1000
ΔV = 0.6
I_C = 0.6 / 1000
I_C = 6 × 10⁻⁴

Q.49. A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 μF is first charged and later connected with resistor 'R'. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be _____________ Ω. [JEE Main 2021]

Ans. 150
T_m =30 ms
C = 200 μF

Since q/N is constant hence
λ =1 / RC

= 150Ω 

Q.50. A certain metallic surface is illuminated by monochromatic radiation of wavelength λ. The stopping potential for photoelectric current for this radiation is 3V0. If the same surface is illuminated with a radiation of wavelength 2λ, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is ____________ λ. [JEE Main 2021]

Ans. 4

Using (i) & (ii)

Q.51. If 2.5 × 10⁻⁶ N average force is exerted by a light wave on a non-reflecting surface of 30 cm2 area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___________ W/cm². (Round off to the Nearest Integer) [JEE Main 2021]
(Assume complete absorption and normal incidence conditions are there)

Ans. 25
Pressure = Intensity / C (for absorbing surface)
I = P × C

I = 25 W/cm²

Q.52. Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is ___________. [JEE Main 2021]

Ans. 1
KE_max = hv − ϕ
1/2mv² = hv − ϕ

Given hv₁ = 2ϕ
hv₂ = 10ϕ
∴

∴ x  = 1

Q.53. In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q and R are E_P, E_Q and E_R, respectively, and they are related by E_P = 2E_Q = 2E_R. In this experiment, the same source of monochromatic light is used for metals P and Q while a different source of monochromatic light is used for the metal R. The work functions for metals P, Q and R are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in eV, is ___________. [JEE Main 2021]

Ans. 6
E_P = 2E_Q = 2E_R = E(assume)
For metal P and Q
E₁ - 4 = E_P = E ……..(1)
E₁ - 4.5 = E_A = E/2 ……..(2)
Subtracting equation (2) from (1), we get
E/2 = 0.5
For metal R
E₂ - 5.5 = E_R = E/2 = 0.5
⇒ E₂ = 6

Q.54. A heavy nucleus N, at rest, undergoes fission N → P + Q, where P and Q are two lighter nuclei. Let δ = M_N − M_P − M_Q, where M_P, M_Q and M_N are the masses of P, Q and N, respectively. EP and EQ are the kinetic energies of P and Q, respectively. The speeds of P and Q are v_P and v_Q, respectively. If c is the speed of light, which of the following statement(s) is(are) correct? [JEE Advance 2021]
(a)
(b)
(c)
(d) The magnitude of momentum for P as well Q is  c2μδ, where

Ans. a, c, d

(p_total) = 0 ⇒ (p_total)f = 0
So, momentum of one nucleus is p in forward direction, then momentum of the other nucleus will be p in backwards direction.
Energy released = (Δm)c² = δc² = KE_P + KE_Q

Q.55. Which of the following statement(s) is(are) correct about the spectrum of the hydrogen atom? [JEE Advance 2021]
(a) The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9/5
(b) There is an overlap between the wavelength ranges of Balmer and Paschen series
(c) The wavelengths of Lyman series are given by (1 + 1/m²)λ₀, where λ₀ is the shortest wavelength of Lyman series and m is an integer
(c) The wavelength ranges of Lyman and Balmer series do not overlap

Ans. a, d