Light of wavelengths 400 nanometer and 8...

Light of wavelengths 400 nanometer and 800 nanometer produces photoemission in two photoemitters. The ratio of the work function of two emitters is

`1:4`

`1:2`

`2:1`

`1:3`

Text Solution

Verified by Experts

The correct Answer is:

`(W_(1))/(W_(2))=(hc)/(lambda_(1))xx(lambda_(2))/(hc)=(lambda_(2))/(lambda_(1))=2:1`

Topper's Solved these Questions

ELECTRONS AND PHOTONS
MARVEL PUBLICATION|Exercise TEST YOUR GRASP - 17|15 Videos
ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS
MARVEL PUBLICATION|Exercise TEST YOUR GRASP - 16|30 Videos
ELECTROSTATICS
MARVEL PUBLICATION|Exercise TEST YOUR GRASP|20 Videos

Similar Questions

Explore conceptually related problems

When light of wavelength 3600 nm falls on a photosensitive plate, photoelectrons are emitted. However, for another photoelectric emitter, light of 6000 nm is sufficient to produce photoemission. What is the ratio of the work functions of the two emitters ?

When light of wavelength 300 nm (nanometre) falls on a photoelectric emitter , however light of 600 nm wavelength is sufficient for creating photoemission . What is the ratio of the work functions of the two emitters ?

When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitted , light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

When light of wavelength 300nm (nanometer) falls on a photoelectric emitter, photoelectrons are liberated. For another emitter,however light of 600nm wavelength is sufficient for creating photoemission. The ratio of the work functions of the two emitters is 2 : x, then the value of x is

When light of wavelength 3000 Å falls on a photosensitive surface (A) photoelectrons are emitted. However, for another photoemitter (B), light of wavelength 6000 Å is required for the emission of photoelectrons. What is the ratio of the work functions of A and B ?

When light of wavelength 300 nm or less falls on aphotoelectric emitter A, photoelectrons are emitted. For another emitter B, light of wavelength 600 nm is sufficient for releasing photoelectorns. The ratio of the work function of emitter A to B is

A light of wavelength 600 nm is incident on a metal surface. When light of wavelength 400 nm is incident, the maximum kinetic energy of the emitted photoelectrons is doubled. The work function of the metals is

MARVEL PUBLICATION-ELECTRONS AND PHOTONS -TEST YOUR GRASP - 17

Light of wavelengths 400 nanometer and 800 nanometer produces photoem...
Text Solution
|
The energy of a photon of wavelength 4000 Å is (use h=6.66xx10^(-34...
Text Solution
|
In a photoelectric experiment, the reciprocal of the slope of the st...
Text Solution
|
The momentum of a photon of an electromagnetic radiation is 3.3xx10^(-...
Text Solution
|
If the threshold frequency for photoemission on a metal corresponds t...
Text Solution
|
Photons of energy 5.5 eV are incident on a metal surface. If the stopp...
Text Solution
|
When light of energy 2.5 eV falls on a metal surface, the maximum K.E....
Text Solution
|
The threshold energy of a surface is 2 eV. When irradiated with some r...
Text Solution
|
Sodium and copper have work functions of 2.3 eV and 4.5 eV. The ratio ...
Text Solution
|
The surface of a metal is illuminated with light of wavelength 400 nm....
Text Solution
|
The current in a photocell can be just reduced to zero by a potential ...
Text Solution
|
A lamp is placed at a distance of 8 cm from a photocell and the microa...
Text Solution
|
The cathode of a photoelectric cell is changed such that the work func...
Text Solution
|
The wavelength of the photons are 4500 Å and 6000 Å. The ratio of thei...
Text Solution
|
An AIR station is broadcasting the waves of wavelength 300 metres. If ...
Text Solution
|
For a photosensitive metal surface the graph of the stopping potential...
Text Solution
|