NCERT Class XII Chapter
Dual Nature of Radiation and Matter
Questions With Solutions
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Question : 30
Total: 37
Light of intensity 10 – 5 W m – 2 falls on a sodium photo-cell of surface area 2 c m 2 . Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wavepicture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Solution:
Wave picture of radiation state that incident energy is uniformly distributed among all the electrons continuously. Let us first calculate the total number of recipient electrons in 5 layers of sodium.
Consider each sodium atom has one electron free as conduction electron.
Effective atomic area =10 – 20 m 2
Number of conduction electrons in 5 layers
n =
or n =
= 10 17
Incident power = incident intensity × area =10 – 5 × 2 × 10 – 4 = 2 × 10 – 9 W
As incident energy is equally distributed among all conduction electrons.
Energy to each conduction electron per second =
= 2 × 10 − 26 W
Time required for emission by each electron
t =
t =
=
= 1.6 × 10 7 s = 0.5 year
where experimental observation shows that emission of photoelectrons is instantaneous ≈10 – 9 sec
Thus, wave picture fails to explain photoelectric effect.
Consider each sodium atom has one electron free as conduction electron.
Effective atomic area =
Number of conduction electrons in 5 layers
n =
Incident power = incident intensity × area =
As incident energy is equally distributed among all conduction electrons.
Energy to each conduction electron per second =
Time required for emission by each electron
t =
t =
where experimental observation shows that emission of photoelectrons is instantaneous ≈
Thus, wave picture fails to explain photoelectric effect.
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