MHT CET · Physics · Dual Nature of Matter
A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum KE of the emitted photoelectrons would be
- A Twice the original value
- B Four times the original value
- C One fourth of the original value
- D Unchanged
Answer & Solution
Correct Answer
(D) Unchanged
Step-by-step Solution
Detailed explanation
The maximum kinetic energy of photoelectrons is given by …. (i)
Where h = Planck’s constant,
v= frequency of radiation
and = threshold frequency.
It can be seen from Eq. (i), that the maximum KE of emitted photoelectrons is proportional to the frequency of radiation and is independent of the intensity of radiation, so it remains unchanged.
Where h = Planck’s constant,
v= frequency of radiation
and = threshold frequency.
It can be seen from Eq. (i), that the maximum KE of emitted photoelectrons is proportional to the frequency of radiation and is independent of the intensity of radiation, so it remains unchanged.
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