Dual Nature of Radiation and Matter - Online Test

Q1. The momentum of photon having energy E is
Answer : Option D
Explaination / Solution:
No Explaination.


Q2. Photo-electric effect can be explained only by assuming that light
Answer : Option B
Explaination / Solution:

Einstein proposed an explanation of the photoelectric effect using a concept first put forward by Max Planck that light waves consist of tiny bundles or packets of energy known as photons or quanta. When a photon falls on the surface of a metal, the entire photon’s energy is transferred to the electron. A part of this energy is used to remove the electron from the metal atom’s grasp and the rest is given to the ejected electron as kinetic energy. Electrons emitted from underneath the metal surface lose some of the kinetic energy in collisions. But the surface electrons carry all the kinetic energy imparted by the photon and have the maximum kinetic energy. 

We can write this mathematically as E = W + Kmax


Q3. The energy of a photon corresponding to the visible light of maximum wavelength is approximately
Answer : Option B
Explaination / Solution:

maximum wavelength of visible light = 7700 A0

Energy


Q4. In order to increase the kinetic energy of ejected photoelectrons, there should be an increase in
Answer : Option D
Explaination / Solution:


Work function depends only on nature of metal. 

So that kinetic energy of an electron increase when frequency of radiation increases.


Q5. If h is Planck’s constant, the momentum of a photon of wavelength 0.01  is
Answer : Option A
Explaination / Solution:
No Explaination.


Q6. Planck’s constant has the dimensions of
Answer : Option B
Explaination / Solution:

Plank Constant and Angular momentum both have same dimension 
Q7. Number of ejected photoelectrons increases with increase
Answer : Option D
Explaination / Solution:
No Explaination.


Q8. A photon behaves as if it had a mass equal to
Answer : Option A
Explaination / Solution:
No Explaination.


Q9.
If the energy of a photon corresponding to a wave length of 6000  is 3.32  joule, the photon energy for a wavelength of 4000  will be

Answer : Option D
Explaination / Solution:

E=hcλE1E2=λ2λ13.32×1019E2=40006000E2=6×3.32×10194=4.98×1019J
Q10. The work function for aluminium surface is 4.2 eV. The cutoff wavelength for the photo electric effect for the surface is
Answer : Option B
Explaination / Solution:

ϕ0=hcλ04.2×1.6×1019=6.6×1034×3×108λ0λ0=6.6×1034×3×1084.2×1.6×1019=2.946×107=2946A