If the energy of incident radiation is increased by 25%. The kinetic energy of the photoelectrons emitted from the metal surface increased from 0.6 eV to 0.9 eV. The initial energy is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.2 eV
According to photoelectric equation,
$K.E.=E_{incident}-\phi$
$⇒K.E._{initial}=E_{initial}-\phi=0.6eV$
and,
$E_{final}=1.25×E_{initial}$
$∴K.E_{final}=E_{final}-\phi=0.9eV$
$⇒E_{initial}-phi=0.6eV$ ...(1)
$1.25×E_{initial}-phi=0.9eV$ ...(2)
Subtracting (1) from (2),
$0.25E_{initial}=0.3eV$
$E_{initial}=\frac{0.3}{0.25}=1.2eV$