Photons of wavelength lamda emitted by a...

Photons of wavelength `lamda` emitted by a source of power P incident on a photo cell. If the current produced in the cell is I, then the percentage of incident photons which produce current in the photo cell is. (where, h is Planck's constant and c is the speed of light in vacuum)

`(100ePc)/(lhlamda)`

`(100ePlamda)/(lhc)`

`(100lhlamda)/(ePc)`

`(100lhc)/(ePlamda)`

Text Solution

Verified by Experts

The correct Answer is:

In photoemission, only one electron can produce one photon.
So, if only x % of incident photons can produce electrons, then
`I=n/t=(100xxIxxhc)/(ePlamda)`

Topper's Solved these Questions

ONLINE QUESTION PAPER
TS EAMCET PREVIOUS YEAR PAPERS|Exercise PHYSICS|40 Videos
QUESTION PAPER 2015
TS EAMCET PREVIOUS YEAR PAPERS|Exercise PHYSICS|40 Videos

Similar Questions

Explore conceptually related problems

Initially a photon of wavelength lambda_(1) falls on the photocathode and emits of maximum energy E_(1) .If the wavelength of the incident photon is changed to lambda_(2) the maximum energy of the electron emitted becomes E_(2) .Then value of he (h=Planck's constant,c= velocity of light) is

Consider the two following A and B and identify the correct choice given in the answers. A In the photovoltaic cells the photoelectric current produced is not proportional to the intensity of incident light . B In gas filled photoemissive cells ,the velocity of photoelectrons depends on the wavelength of the incident radiation.

A monochromatic light of wavelength lamda is incident on an isolated metallic sphere of radius a. The threshold wavelength is lamda_(0) , which is larger than lamda . Find the number of photoelectrons emitted before the emission of photo electrons stops.

Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V , estimate the work function of the metal. How would the photo-cell respond to a high intensity (~10^(5) W m^(2)) red light of wavelength 6328 Å produced by a He-Ne laser ?

Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth orbits from higher energy orbits respectively (as shown in figure) Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2) . For example, in the case of n = 4, number of lines produced is 6. (4 rarr 3, 4 rarr 2, 4 rarr 1, 3 rarr 2, 3 rarr 1, 2 rarr 1) . When an electron returns from n_(2) to n_(1) state, the number of lines in the spectrum will be equal to ((n_(2) - n_(1))(n_(2)-n_(1) +1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(2) then the difference may be expressed in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = (h c)/(Delta E) . Since h and c are constant, Delta E corresponds to definite energy, thus each transition from one energy level to another will prouce a higher of definite wavelength. THis is actually observed as a line in the spectrum of hydrogen atom. Wave number of the line is given by the formula bar(v) = RZ^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Where R is a Rydberg constant (R = 1.1 xx 10^(7)) (i) First line of a series : it is called .line of logest wavelength. or .line of shortest energy.. (ii) Series limit of last of a series : It is the line of shortest wavelength or line of highest energy. The difference in the wavelength of the 2^(nd) line of Lyman series and last line of Bracket series in a hydrogen sample is

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth orbits from higher energy orbits respectively (as shown in figure) Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2) . For example, in the case of n = 4, number of lines produced is 6. (4 rarr 3, 4 rarr 2, 4 rarr 1, 3 rarr 2, 3 rarr 1, 2 rarr 1) . When an electron returns from n_(2) to n_(1) state, the number of lines in the spectrum will be equal to ((n_(2) - n_(1))(n_(2)-n_(1) +1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(2) then the difference may be expressed in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = (h c)/(Delta E) . Since h and c are constant, Delta E corresponds to definite energy, thus each transition from one energy level to another will prouce a higher of definite wavelength. THis is actually observed as a line in the spectrum of hydrogen atom. Wave number of the line is given by the formula bar(v) = RZ^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Where R is a Rydberg constant (R = 1.1 xx 10^(7)) (i) First line of a series : it is called .line of logest wavelength. or .line of shortest energy.. (ii) Series limit of last of a series : It is the line of shortest wavelength or line of highest energy. In a single isolated atom in electron makes transition from 5^(th) excited state to 2^(nd) state the maximum number of different types of photons observed as

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth orbits from higher energy orbits respectively (as shown in figure) Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2) . For example, in the case of n = 4, number of lines produced is 6. (4 rarr 3, 4 rarr 2, 4 rarr 1, 3 rarr 2, 3 rarr 1, 2 rarr 1) . When an electron returns from n_(2) to n_(1) state, the number of lines in the spectrum will be equal to ((n_(2) - n_(1))(n_(2)-n_(1) +1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(2) then the difference may be expressed in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = (h c)/(Delta E) . Since h and c are constant, Delta E corresponds to definite energy, thus each transition from one energy level to another will prouce a higher of definite wavelength. THis is actually observed as a line in the spectrum of hydrogen atom. Wave number of the line is given by the formula bar(v) = RZ^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Where R is a Rydberg constant (R = 1.1 xx 10^(7)) (i) First line of a series : it is called .line of logest wavelength. or .line of shortest energy.. (ii) Series limit of last of a series : It is the line of shortest wavelength or line of highest energy. The wave number of electromagnetic radiation emitted during the transition of electron in between two levels of Li^(2+) ion whose principal quantum numbers sum if 4 and difference is 2 is

TS EAMCET PREVIOUS YEAR PAPERS-ONLINE QUESTION PAPER 2018-PHYSICS

A source S emitting sound of frequency 288 Hz is fixed on block B whic...
Text Solution
|
In a compound microscope, the focal lengths of two lenses are 1.5 cm a...
Text Solution
|
In Young's double slit experiment of central fringe is I(0) and fringe...
Text Solution
|
A proton and an alpha-particle start from rest in a uniform electric f...
Text Solution
|
Two charged balls moving in the same direction with same velocity v ar...
Text Solution
|
Electric field vector in a region is given by E=(3hati+4hatj)V-m^(-1)....
Text Solution
|
In the circuit shown in figure, if the point R is earthed and point P ...
Text Solution
|
The bulb which glows with maximum intensity in the given circuit is
Text Solution
|
In the circuit shown in figure, power developed across 1Omega,2Omegaan...
Text Solution
|
Two long straight parallel conductors are carrying currents i(1)andi(2...
Text Solution
|
A straight conductor of length 32 cm carries a current of 30 A. Magnet...
Text Solution
|
A sample of a paramagnetic salt containing 3xx10^(24) atomic dipoles e...
Text Solution
|
A coil of wire of radius r has 600 turns and self inductance of 108 mH...
Text Solution
|
In the AC circuit shown, E=E(0)sin(omegat+phi)andi=i(0)sin(omegat+ph...
Text Solution
|
The oscillating electric field of an electromagnetic wave is given by ...
Text Solution
|
Photons of wavelength lamda emitted by a source of power P incident on...
Text Solution
|
If lamda(1)andlamda(2) are the wavelength of the photons emitted, when...
Text Solution
|
The number of half lives elapsed before 93.75% of a radioactive sample...
Text Solution
|
In the common-base configuration, a transistor has current amplificati...
Text Solution
|
If the height of the transmitting tower is increased by 30%, then the ...
Text Solution
|