The document NEET Previous Year Questions (2014-20): Dual Nature of Radiation and Matter Notes | EduRev is a part of the Class 12 Course Physics 28 Years Past year papers for NEET/AIPMT Class 12.

All you need of Class 12 at this link: Class 12

**Q.1. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled? (2020)A: one-fourth B: zero **

D: four times

Solution:

f

∴ f

for given condition

f

so no photo electron emission

i = 0

A: 12.2 × 10

B: 12.2 × 10

C: 12.2 × 10

D: 12.2 nm

Ans: B

Solution:

For an electron accelerated through a potential V

**Q.3. An electron of mass m with an initial velocityenters an electric field E _{0} = constant > 0) at t = 0. If λ_{0} is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is :- (2018)**

A:

B:

**C: λ _{0}tD: λ_{0}Ans: **A

Solution:

A: 1 : 2

B: 1 : 4

C: 4 : 1

D: 2 : 1

Ans:

(Given h = 4.14 × 10

A: ≈ 0.6 × 10

B: ≈ 61 × 10

C: ≈ 0.3 × 10

D: ≈ 6 × 10

Solution:

A: 3λ

B: 4λ

C: 5λ

D: 5/2λ

Ans:

When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V.

Photoelectric equation can be written as,

Now, when the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. So, photoelectric equation can be written as,

From equations (i) and (ii), we get

When a metallic surface is illuminated with radiation of wavelength λ stopping potential is V.

Photoelectric equation can be written as,

Now, when the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. So, photoelectric equation can be written as,

From equations (i) and (ii), we get

A:

B:

C:

**D: c(2mE) ^{1/2}Ans: **B

Solution:

Given that electron has a mass m.

De-Broglie wavelength for an electron will be given as,

where,

h is the Planck's constant, and

p is the linear momentum of electron

Kinetic energy of electron is given by,

From equation (i) and (ii), we have

Energy of a photon can be given as,

Hence, λ

Now, dividing equation (iii) by (iv), we get

A: λ/6

B: 6λ

C: 4λ

D: λ/4

Ans:

We have,

where W is the work function and (3V_{0}) is the stopping potential when monochromatic light of wavelength λ is used.

where V_{0} is the stopping potential when monochromatic light of wavelength 2λ is used.

Subtracting equation (2) from equation (1)

We get,

∴

Substituting in equation (2) we get,

∴

The threshold wavelength is therefore 4λ.**Q.9.** **Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength? (2015)A:**

B:

C:

D:

**Ans: **C

Solution:

The de-Broglie wavelength is given by

This equation is in the form of yx = c, which is the equation of a rectangular hyperbola.

The de-Broglie wavelength is given by

This equation is in the form of yx = c, which is the equation of a rectangular hyperbola.

**Q.10. ****When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is : (2014)A: 1.3 eVB: 1.5 eVC: 0.65 eVD: 1.0 eVAns: **D

Original energy of photon be E

From equation (i) and (ii)

A: 60

B: 50

C: 25

D: 75

Ans:

Solution:

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

14 docs|25 tests