The document Previous year questions (2016-20): Modern Physics- 1 | JEE Main Notes | EduRev is a part of the JEE Course Class 12 Physics 35 Years JEE Mains &Advance Past year Paper.

All you need of JEE at this link: JEE

**Q.1. A beam of electromagnetic radiation of intensity 6.4 x 10 ^{-5} W/cm^{2 }is comprised of wavelength, λ = 310 nm. It falls normally on a metal (work function ϕ = 2eV) of surface area 1 cm^{2 }. If one in 10^{3} protons ejects an electron, total number of electrons ejected in 1s is 10^{x},(hc = 1240 eV nm, 1eV 1.6 10^{-19} J) then x is ________. (2020)** (11)

Ans.

Given that λ = 310 nm, hc = 1240 eV nm We know that photons will eject electrons if the energy of incident beam is greater than work function, that is, E > ϕ

⇒ hf > ϕ ⇒ hc/λ > ϕ

⇒

⇒ 4 eV > 2 eV

So, electrons will be ejected.

Now number of electrons ejected in 1s is

n

Number of photon incident is given by

So,

Therefore, x = 11

(1)

(2)

(3)

(4)

Ans.

Solution.

Given that

de-Broglie wavelength before entering in electric filed is given by

Velocity of electron after entering electric filed at time t is given by

Since, electric filed is at z direction so it effects only z direction velocity.

So, de-Broglie wavelength after entering in electric filed at time t is given by

= **Q.3. Radiation with wavelength 6561 Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10 ^{−4} T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to (2020)(1) 1.1 eV(2) 0.8 eV(3) 1.6 eV(4) 1.8 eVAns. **(1)

Solution.

Given that

Einstein’s Photoelectric Equation is given as

............ (1)

Radius of circular path of charged particle is magnetic field is given by

For maximum radius K should be maximum. So,

............. (2)

From Eqs. (1) and (2), we get

=

= 1.1eV**Q.4. A particle moving with kinetic energy E has de Broglie wavelength λ. If energy ΔE is added to its energy, the wavelength become λ/2. Value of ΔE, is (2020)(1) E(2) 4E(3) 3E(4) 2EAns. **(3)

Solution.

de Broglie wavelength is given by

So,

**Q.5. A electron of mass m and magnitude of charge |e| initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is (2020)****(a)****(b)****(c)****(d)****Ans. **(4)**Solution.**

Electrostatic Force on a charge particle in a electric filed is given by

F = |e| E

Initially particle is at rest, that is, u = 0. So, velocity of particle after time t in electric filed is

De-Broglie wavelength is given by

So, the rate of change of de-Broglie wavelength of this electron at time t is given by**Q.6. The activity of a radioactive sample falls from 700/s to 500/s in 30 mins. Its half-life is close to (2020)(1) 72 min(2) 62 min(3) 66 min(4) 52 minAns.** (2)

Given that, A

We know that, radioactivity at instant t is given by

We know that, half-life is given by

Q.7. The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10^{−16}s. The frequency of revolution of the electron in its first excited state (in/s) is (2020)

(1) 1.6 × 10^{14}**(2) 7.8 × 10 ^{14}(3) 6.2 × 10^{15}(4) 5.6 × 10^{12}Ans. **(2)

Solution.

Given that T =

Time period of revolution of n

here

So,

For hydrogen atom Z = 1, T ∝ n

We have n = 1, T = 1.6 × 10

For n = 2, we have

Therefore, frequency is

[θ: Scattering angle

Y: Number of scattered α-particles detected]

(Plots are schematic and not to scale) (2020)

(1)

(2)

(3)

(4)

Relation between scattering angle (θ) and number of scattered α-particle (Y) detected is given by

OR

So, plot of option (4) depicts the correct relation between scattering angle and number of scattered α particle detected.

Ans.

Solution.

Given that

For Balmer series, wavelength is given by

For first member, we have

................. (i)

For second member, we have

................... (ii)

From Eqs. (1) and (2), we have

= 486nm

(1) 24.2 nm

(2) 11.4 nm

(3) 35.8 nm

(4) 8.6 nm

Ans.

Solution.

Given that E

We know that 1Rydberg = 13.6eV

Ionization energy is given by E

So, 13.6Z

Z

Z = 3

The wavelength of the radiation emitted when the electron in this ion jumps from the

second excited state to the ground state is given by

(1) 1.8

(2) 2.5

(3) 5.6

(4) 1.4

Solution.

Now, dividing Eq. (1) by Eq. (2), we get

(c = 3 × 10

(1) 6.82 eV

(2) 12.5 eV

(3) 8.52 eV

(4) 7.72 eV

Ans.

B = B

where c is the speed of light

In above wave equation, there are two electromagnetic waves with different frequency.

To get maximum kinetic energy consider the photon with higher frequency

B

B

v

v

Kinetic energy of photoelectron will be a maximum for photon of higher energy.

E

= 6.6 × 3 × 10

(1) 500 keV

(2) 100 keV

(3) 1 keV

(4) 25 keV

Ans.

Therefore, required energy is

(1) 0.5 V

(2) 1.5 V

(3) 1.0 V

(4) 2.0 V

Ans.

λ

λ

For λ

Subtracting Eq. (2) from Eq. (1), we get

(1) 1700 nm

(2) 2020 nm

(3) 220 nm

(4) 250 nm

Ans.

(3) 2v

Ans.

From Eq. (1) and Eq. (2), we get

(1) 25 a

(2) 9 a

(3) 4 a

(4) 16 a

Ans.

Since, electron will excite to n = 4

(2) 1.1 × 10

(3) 1.7 × 10

(4) 1.8 × 10

Ans.

Ans.

From N-shell to L-shell

Dividing Eq. (1) by Eq. (2), we get

Ans.

In circular motion

Bohr quantization is

From Eq. (1), we get

From Eq. (1), we get

(1) 10.00

Ans.

(1) 5 days and 10 days.

(2) 10 days and 40 days.

(3) 20 days and 5 days.

(4) 20 days and 10 days.

Ans.

A = λN

for A λ

for B λ

Dividing Eq. (1) by Eq. (2), we get

(T

(T

(1) 4ln2

(4) 2ln2

Ans.

N = N

If T

Half-life of A = ln 2 [ln = log

λ

At t = 0 R

(1) 200

(2) 150

(3) 400

(4) 360

Ans.

(1) Energy of 12.4 MeV will be supplied.

(2) 8.3 MeV energy will be released.

(3) Energy of 3.6 MeV will be released.

(4) Energy of 11.9 MeV has to be supplied.

Ans.

Ne

Amount of energy

ΔE = 2 × (Binding energy of He

= 2 × (4 × 7.07) + (12 × 7.86) – (20 × 8.03)

= 56.56 + 94.32 – 160.6

= −9.72 MeV

Hence, the energy of 11.9 MeV has to be supplied.

(1) A = 208; Z = 80

(2) A = 202; Z = 80

(3) A = 208; Z = 82

(4) A = 200; Z = 81

Ans.

Hence Z = 82, A = 208 and element is Pb.

(1)

(2)

(3)

(4)

Ans:

2πr = nλ

From equation (1) and (2)

**(4) Ans:** (4)

Series limit frequency of the Lyman series is given by

Series limit frequency of the Pfund series

(1) (.89, .28)

(2) (.28, .89)

(3) (0, 0)

(4) (0, 1)

Ans:

(from conservation of momentum)

and K

So after solving

**(3) **

**(4) ****Ans:** (3)**Solution:**

Momentum of each electron

Velocity of centre of mass

velocity of 1st particle about centre of mass

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

42 docs|19 tests

- Test: Single Correct MCQs: Modern Physics | JEE Advanced
- Previous year questions (2016-20): Modern Physics- 2 | JEE Main
- Test: MCQs (One or More Correct Option): Modern Physics | JEE Advanced
- Test: 35 Year JEE Previous Year Questions: Modern Physics
- Fill in the Blanks: Modern Physics | JEE Advanced
- True/False: Modern Physics | JEE Advanced